Class 9 Advance Maths(Elective) Chapter 4 Exercise 4.2 (part 1)

4. SPECIAL PRODUCT AND FACTORIZTION

Exercise 4.2

1. Prove that:

(i) (x+y+z)³=(y+z-x)³+(z+x-y)³+(x+y-z)³+24xyz
Solution: Let, y+z-x=a
             z+x-y=b
             x+y-z=c
Therefore,
a+b+c=y+z-x+z+x-y+x+y-z
=x+y+z
a+b=y+z-x+z+x-y
   =2z
b+c=z+x-y+x+y-z
   =2x
c+a=x+y-z+y+z-x
   =2y
Now,
RHS=(y+z-x)³+(z+x-y)³+(x+y-z)³+24xyz
   =a³+b³+c³+3.2x.2y.2z
   =(a+b+c)³
   =(x+y+z)³
   =LHS
Hence, Proved.

(ii) 27(a+b+c)³-(2a+b)³-(2b+c)³-(2c+a)³=3(a+2b+3c)(b+2c+3a)(c+2a+3b)
Solution: Let,  2a+b=x
        2b+c=y
2c+a=z
Therefore,
x+y+z=2a+b+2b+c+2c+a
     =3a+3b+3c
     =3(a+b+c) 
x+y=2a+b+2b+c
   =2a+3b+c
   =c+2a+3b
y+z=2b+c+2c+a
   =2b+3c+a
   =a+2b+3c
z+x=2c+a+2a+b
   =2c+3a+b
   =b+2c+3a
Now,
LHS=27(a+b+c)³-(2a+b)³-(2b+c)³-(2c+a)³
   ={3(a+b+c)}³-x³-y³-z³
   =(x+y+z)³-(x³+y³+z³)
   =3(x+y)(y+z)(z+x)
   =3(c+2a+3b)(a+2b+3c)(b+2c+3a)
   =RHS
Hence, Proved.

(iii) (2a-b-c)³+(2b-c-a)³+(2c-a)³+3(a+b-2c)(2b+c-2a)(c+a-b)=b³
Solution: Let,  2a-b-c=x
2b-c-a=y
2c-a=z
Therefore, x+y+z=2a-b-c+2b-c-a+2c-a
         =b
x+y=2a-b-c+2b-c-a
   =a+b-2c
y+z=2b-c-a+2c-a
   =2b+c-2a
z+x=2c-a+2a-b-c
   =c+a-b
Now,
LHS=(2a-b-c)³+(2b-c-a)³+(2c-a)³+3(a+b-2c)(2b+c-2a)(c+a-b)
   =x³+y³+z³+3(x+y)(y+z)(z+x)
   =(x+y+z)³
   =b³
   =RHS
Hence, Proved.

(iv) (a+2b)³+(b+2c)³+(c+2a)³+3(a+3b+2c)(b+3c+2a)(c+3a+2b)=27(a+b+c)³
Solution: Let,  a+2b=x
b+2c=y
c+2a=z
Therefore,
x+y+z=a+2b+b+2c+c+2a
     =3a+3b+3c
     =3(a+b+c)
x+y=a+2b+b+2c
   =a+3b+2c
y+z=b+2c+c+2a
   =b+3c+2a
z+x=c+2a+a+2b
   =c+3a+2b
Now,
LHS=(a+2b)³+(b+2c)³+(c+2a)³+3(a+3b+2c)(b+3c+2a)(c+3a+2b)
   =x³+y³+z³+3(x+y)(y+z)(z+x)
   =(x+y+z)³
   ={3(a+b+c)}³
   =27(a+b+c)³
   =RHS
Hence, Proved.

(v) (x-2y)³+(2y-3z)³+(3z-x)³=3(x-2y)(2y-3z)(3z-x)
Solution: Let,  x-2y=a
2y-3z=b
3z-x=c
Therefore,
a+b+c=x-2y+2y-3z+3z-x
=>a+b+c=0 [If a+b+c=0,then a³+b³+c³=3abc]
=>a³+b³+c³=3abc
=>(x-2y)³+(2y-3z)³+(3z-x)³=3(x-2y)(2y-3z)(3z-x)
Hence,Proved.

(vi) 8a³-3(a-b)(a+c)(2a+b-c)=(a+b)³+(a-c)³+(c-b)³
Solution: Let,  a+b=x
a-c=y
c-b=z
Therefore,
x+y+z=a+b+a-c+c-b
     =2a
x+y=a+b+a-c
   =2a+b-c
y+z=a-c+c-b
   =a-b
z+x=c-b+a+b 
   =a+c
Now,
RHS=(a+b)³+(a-c)³+(c-b)³
   =x³+y³+z³
   =(x+y+z)³-3(x+y)(y+z)(z+x)
   =(2a)³-3(2a+b-c)(a-b)(a+c)
   =8a³-3(a-b)(a+c)(2a+b-c)
   =LHS
Hence, Proved.

(vii) (2a+b+c)³+(a+2b+c)³+(a+b+2c)³=64(a+b+c)³-3(3a+3b+2c)(3a+2b+3c)(2a+3b+3c)
Solution: Let,  2a+b+c=x
a+2b+c=y
a+b+2c=z
Therefore,
x+y+z=2a+b+c+a+2b+c+a+b+2c
     =4a+4b+4c
     =4(a+b+c)
x+y=2a+b+c+a+2b+c
   =3a+3b+2c
y+z=a+2b+c+a+b+2c
   =2a+3b+3c
z+x=a+b+2c+2a+b+c
   =3a+2b+3c
Now,
LHS=(2a+b+c)³+(a+2b+c)³+(a+b+2c)³
   =x³+y³+z³
   =(x+y+z)³+3(x+y)(y+z)(z+x)
   ={4(a+b+c)}³+3(3a+3b+2c)(2a+3b+3c)(3a+2b+3c)
   =64(a+b+c)³+3(3a+3b+2c)(2a+3b+3c)(3a+2b+3c)
   =RHS
Hence, Proved.


2. Find the product of-

(i) (x+a)(x+3a)(x+5a)
= x³+(a+3a+5a)x²+(a.3a+3a.5a+5a.a)x+a.3a.5a
= x³+9ax²+(3a²+15a²+5a²)x+15a³
= x³+9ax²+3a²x+15a²x+5a²x+15a³
= x³+9ax²+23a²x+15a³

(ii) (x+1)(x+2)(x+3)
= x³+(1+2+3)x²+(1*2+2*3+3*1)x+1*2*3
= x³+6x²+11x+6

(iii) (x-c)(x-d)(x+k)
= x³+(-c-d+k)x²+{(-c).(-d)+(-d).k+k.(-c)}x+(-c).(-d).k
= x³+(-c-d+k)x²+(cd-dk-kc)x+cdk
= x³-cx²-dx²+kx²+cdx-dkx-ckx+cdk

(iv) (a+b+1)(a²+b²+1-ab-b-a)
= (a+b+1)(a²+b²+1²-a.b-b.1-1.a)
= a³+b³+1³-3.a.b.1
= a³+b³+1-3ab

(v) (a+b-c)(a²+b²+c²-ab+bc+ca)
= {a+b+(-c)}{a²+b²+(-c²)-ab-b(-c)-(-c)a}
= a³+b³+(-c)³-3.a.b.(-c)
= a³+b³-c³+3abc

(vi) (x+2y+3z)(x²+4y²+9z²-2xy-6yz-3zx)
= (x+2y+3z){x²+(2y)²+(3z)²-x.2y-2y.3z-3z.x}
= x³+(2y)³+(3z)³-3.x.2y.3z
= x³+8y³+9z³-18xyz

(vii) (3a-5b-2)(9a²+25b²+15ab+6a-10b+4)
= {3a+(-5b)+(-2)}{(3a)²+(-5b)²+(-2)²-3a.(-5b)-(-5b).(-2)-(-2).3a}
= (3a)³+(-5b)³+(-2)³-3.3a.(-5b).(-2)
= 27a³-125b³-8-9a.10b
= 27a³-125b³-8-90ab


3.Divide-






4.Express into factors-

(i) 8x³-(x-y)³-(y-z)³-(z+x)³
Solution:
Let, x-y=a
     y-a=b
     z+x=c
Now, a+b+c=x-y+y-a+z+x
          =2x
Therefore the given expression,
(2x)³-(x-y)³-(y-z)³-(z+x)³
= (a+b+c)³-a³-b³-c³
= 3(a+b)(b+c)(c+a)
= 3(x-y+y-z)(y-z+z+x)(z+x+x-y)
= 3(x-z)(x+y)(2x-y+z)
= 3(x+y)(x-z)(2x-y+z)

(ii) x⁶+5x³+8
Solution:
  x⁶+5x³+8
= (x²)³-x³+6x³+2³
= (x²)³+(-x)³+2³-3.x².(-x).2
= (x²-x+2)[(x²)²+(-x)²+2²-x².(-x)-(-x).2-2.x²]
= (x²-x+2)(x⁴+x²+4+x³+2x-2x²)

(iii) 8(x+y+z)³-(y+z)³-(z+x)³-(x+y)³
Solution:
Given,   8(x+y+z)³-(y+z)³-(z+x)³-(x+y)³
       = {2(x+y+z)}³-(y+z)³-(z+x)³-(x+y)³
Let y+z=a
    z+x=b
    x+y=c
Now,
a+b+c=y+z+z+x+x+y
     =2(x+y+z)
a+b=y+z+z+x
   =x+y+2z
b+c=z+x+x+y
   =2x+y+z
c+a=x+y+y+z
   =x+2y+z
Therefore the given expression,
{2(x+y+z)}³-(y+z)³-(z+x)³-(x+y)³
= (a+b+c)³-a³-b³-c³
= 3(a+b)(b+c)(c+a)
= 3(x+y+2z)(2x+y+z)(x+2y+z)

(iv) x³-(a+b+c)x²+(ab+bc+ca)x-abc
Solution:
x³-(a+b+c)x²+(ab+bc+ca)x-abc
= x³-ax²-bx²-cx²+abx+bcx+cax-abc
= x²(x-a)-bx(x-a)-cx(x-a)+bc(x-a)
= (x-a)(x²-bx-cx+bc)
= (x-a)[x(x-b)-c(x-b)]
= (x-a)(x-b)(x-c)

(v) t³+(a+b-c)t²-(bc+ca-ab)t-abc
Solution:
t³+(a+b-c)t²-(bc+ca-ab)t-abc
= t³+at²+bt²-ct²-bct-cat+abt-abc
= t²(t+a)+bt(t+a)-ct(t+a)-bc(t+a)
= (t+a)(t²+bt-ct-bc)
= (t+a)[t(t+b)-c(t+b)]
= (t+a)(t+b)(t-c)

(vi)(x+3y+12z)(12yz+4zx+xy)-12xyz
Solution:
(x+3y+12z)(12yz+4zx+xy)-12xyz
= 1/3*3[(x+3y+12z)(12yz+4zx+xy)-12xyz]
= 1/3[3(x+3y+12z)(12yz+4zx+xy)-3*12xyz]
= 1/3[(x+3y+12z)(36yz+12zx+3xy)-36xyz]
= 1/3[(x+3y+12z)(3y.12z+12z.x+x.3y)-x.3y.12z]
= 1/3(x+3y)(3y+12z)(12z+x)
= 1/3*3(x+3y)(y+6z)(12z+x)
= (x+3y)(y+6z)(12z+x)

(vii) a²(3c-2b)+4b²(a+3c)+9c²(a-2b)-12abc
Solution:
a²(3c-2b)+4b²(a+3c)+9c²(a-2b)-12abc
= a²{3c+(-2b)}+(-2b)²(a+3c)+(3c)²{a+(-2b)}-2.a.(-2b).3c
= (3c-2b)(a+3c)(a-2b)

(viii) 2x²y²z²+x²y²(x²+y²)+y²z²(y²+z²)+z²x²(z²+x²)
Solution:
2x²y²z²+x²y²(x²+y²)+y²z²(y²+z²)+z²x²(z²+x²)
= 2.x².y².z²+x².y²(x²+y²)+y².z²(y²+z²)+z².x²(z²+x²)
= (x²+y²)(y²+z²)(z²+x²)

(ix) a(b-c)²+b(c-a)²+c(a-b)²+8abc
Solution:
a(b-c)²+b(c-a)²+c(a-b)²+8abc
= a(b²-2bc+c²)+b(c²-2ca+a²)+c(a²-2ab+b²)+8abc
= ab²-2abc+c²a+bc²-2abc+a²b+ca²-2abc+b²c+8abc
= ab²+c²a+bc²+a²b+ca²+b²c+2abc
= ab²+c²a+bc²+a²b+ca²+b²c+abc+abc
= (ab²+a²b)+(c²a+bc²)+(ca²+abc)+(b²c+abc)
= ab(b+a)+c²(b+a)+ac(b+a)+bc(b+a)
= (b+a)(ab+c²+ac+bc)
= (b+a)(ab+ac+c²+bc)
= (a+b)[a(b+c)+c(c+b)]
= (a+b)(b+c)(c+a)

(x) x(y²+z²)-y(z²+x²)-z(x²+y²)+3xyz
Solution:
x(y²+z²)-y(z²+x²)-z(x²+y²)+3xyz
= x{(-y²)+(-z²)}+(-y){(-z²)+x²)}+(-z){x²+(-y²)}+3.x.(-y).(-z)
= {x+(-y)+(-z)}{x.(-y)+(-y).(-z)+(-z).x}
=(x-y-z)(yz-xy-zx)

(xi) 3xy(y-3z)+2yz(y-2z)+6xz(3x+2z)-18xyz
Solution:
3xy(y-3z)+2yz(y-2z)+6xz(3x+2z)-18xyz
= 3x.(-y){(-y)+3z} + 2z.(-y){(-y)+2z)} + 3x.2z(3x+2z) + 3.3x.(-y).2z
= {3x+(-y)+2z}{3x.(-y)+(-y).2z+2z.3x}
= (3x-y+2z)(6xz-3xy-2yz)

(xii) a(b+c)²+b(c+a)²+c(a+b)²-4abc
Solution:
a(b+c)²+b(c+a)²+c(a+b)²-4abc
= a(b²+2bc+c²)+b(c²+2ca+a²)+c(a²+2ab+b²)-4abc
= ab²+2abc+c²a+bc²+2abc+a²b+ca²+2abc+b²c-4abc
= ab²+c²a+bc²+a²b+ca²+b²c+2abc
= ab²+c²a+bc²+a²b+ca²+b²c+abc+abc
= ab²+a²b+c²a+bc²+abc+ca²+b²c+abc
= ab(b+a)+c²(a+b)+ac(b+a)+bc(b+a)
= (a+b)(ab+c²+ca+bc)
= (a+b)(ab+bc+c²+ca)
= (a+b){b(a+c)+c(c+a)}
= (a+b)(b+c)(c+a)

(xiii) (y+z)²(2x+y+z)+(z+x)²(x+2y+z)+(x+y)²(x+y+2z)+3(x+y)(y+z)(z+x)
Solution:
(y+z)²(2x+y+z)+(z+x)²(x+2y+z)+(x+y)²(x+y+2z)+3(x+y)(y+z)(z+x)
= (y+z)²[(x+y)+(z+x)]+(z+x)²[(x+y)+(y+z)]+(x+y)²[(z+x)+(y+z)]+3(x+y)(y+z)(z+x)
= [(y+z)+(z+x)+(x+y)][(y+z)(z+x)+(z+x)(x+y)+(x+y)(y+z)]
= (y+z+z+x+x+y)(yz+xy+z²+zx+zx+yz+x²+xy+xy+zx+y²+yz)
= 2(x+y+z)(x²+y²+z²+3xy+3yz+3zx)

(xiv) (x+y)²(x²+y²+x-y)+(x-y)²(x²+y²+x+y)+2x(x²+y²)²+3(x⁴-y⁴)
Solution:
Let, x+y=a
     x-y=b
     x²+y²=c
Now, 
a+b+c=x+y+x-y+x²+y²
     =2x+x²+y²
a+b=x+y+x-y
   =2x
b+c=x-y+x²+y²
   =x²+y²+x-y
c+a=x²+y²+x+y
a.b.b=(x+y)(x-y)(x²+y²)
     =(x²-y²)(x²+y²)
     =x⁴-y⁴
Therefore the given expression,
(x+y)²(x²+y²+x-y)+(x-y)²(x²+y²+x+y)+2x(x²+y²)²+3(x⁴-y⁴)
= a²(b+c)+b²(c+a)+c²(a+b)+3abc
= (a+b+c)(ab+bc+ca)
= (2x+x²+y²)[(x+y)(x-y)+(x-y)(x²+y²)+(x²+y²)(x+y)]
= (2x+x²+y²)(x²-y²+x³+xy²-x²y-y³+x³+x²y+xy²+y³)
= (2x+x²+y²)(x²-y²+2x³+2xy²)

(xv) 2(a²b+b²c+c²a)+4(ab²+bc²+ca²)+9abc
Solution:
2(a²b+b²c+c²a)+4(ab²+bc²+ca²)+9abc
= 2a²b+2b²c+2c²a+4ab²+4bc²+4ca²+8abc+abc
= (2b²+c+2c²a+4bc²+abc)+(2a²b+4ca²+4ab²+8abc)
= c(2b²+2ca+4bc+ab)+2a(ab+2ca+2b²+4bc)
= c(ab+2ca+2b²+4bc)+2a(ab+2ca+2b²+4bc)
= (ab+2ca+2b²+4bc)(c+2a)
= (c+2a)[(ab+2b²)+(2ca+4bc)]
= (c+2a)[b(a+2b)+2c(a+2b)]
= (c+2a)(a+2b)(b+2c)

(xvi) (a+b)(ab+c²)+c(a²+b²)+3abc
Solution:
(a+b)(ab+c²)+c(a²+b²)+3abc
= a(ab+c²)+b(ab+c²)+c(a²+b²)+3abc
= a²b+c²a+ab²+bc²+ca²+b²c+3abc
= ab²+c²a+a²b+bc²+ca²+b²c+3abc
= a(b²+c²)+b(c²+a²)+c(a²+b²)+3abc
= (a+b+c)(ab+bc+ca)

(xvii) x(y-z)²+y(z-x)²+z(x-y)²+8xyz
Solution:
x(y-z)²+y(z-x)²+z(x-y)²+8xyz
= x(y²-2yz+z²)+y(z²-2zx+x²)+z(x²-2xy+y²)+8xyz
= xy²-2xyz+z²x+yz²-2xyz+x²y+zx²-2xyz+y²z+8xyz
= xy²+z²x+yz²+x²y+zx²+y²z+2xyz
= x²y+xy²+y²z+yz²+z²x+zx²+2xyz
= xy(x+y)+yz(y+z)+zx(z+x)+2xyz
= (x+y)(y+z)(z+x)

(xviii) (x²-yz)³+(y²-xz)³+(z²-xy)³-3(x²-yz)(y²-xz)(z²-xy)
Solution:
Let, x²-yz=a
     y²-xz=b
     z²-xy=c
a+b+c= x²-yz+y²-xz+z²-xy
     = x²+y²+z²-xy-yz-xz
a-b= x²-yz-y²+xz
   = x²-y²+z(x-y)
   = (x-y)(x+y)+z(x-y)
   = (x-y)(x+y+z)
b-c= y²-xz-z²+xy
   = y²-z²+x(y-z)
   = (y+z)(y-z)+x(y-z)
   = (y-z)(x+y+z)
c-a= z²-xy-x²+yz
   = z²-x²+y(z-x)
   = (z+x)(z-x)+y(z-x)
   = (z-x)(x+y+z)
Therefore the given expression,
(x²-yz)³+(y²-xz)³+(z²-xy)³-3(x²-yz)(y²-xz)(z²-xy)
= a³+b³+c³-3abc
= (a+b+c)(a²+b²+c²-ab-bc-ca)
= 1/2*2(a+b+c)(a²+b²+c²-ab-bc-ca)
= 1/2(a+b+c)(2a²+2b²+2c²-2ab-2bc-2ca)
= 1/2(a+b+c)[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)]
= 1/2(a+b+c)[(a-b)²+(b-c)²+(c-a)²]
= 1/2(a+b+c)[{(x-y)(x+y+z)}²+{(y-z)(x+y+z)}²+{(z-x)(x+y+z)}²]
= 1/2(a+b+c){(x-y)²(x+y+z)²+(y-z)²(x+y+z)²+(z-x)²(x+y+z)²}
= 1/2(a+b+c)[(x+y+z)²{(x-y)²+(y-z)²+(z-x)²}]
= 1/2(x²+y²+z²-xy-yz-xz)(x+y+z)²(x²-2xy+y²+y²-2yz+z²+z²-2zx+x²)
= 1/2(x+y+z)²(x²+y²+z²-xy-yz-xz)(2x²+2y²+2z²-2xy-2yz-2zx)
= 1/2*2(x+y+z)²(x²+y²+z²-xy-yz-xz)(x²+y²+z²-xy-yz-xz)
= (x+y+z)²(x²+y²+z²-xy-yz-xz)²
= [(x+y+z)(x²+y²+z²-xy-yz-xz)]²
= (x³+y³+z³-3xyz)²

(xix) (5l-2m)³+(m-2n)³+(m+2n-5l)³
Solution:
Let, 5l-2m=a
     m-2n=b
     m+2n-5l=c
a+b+c=5l-2m+m-2n+m+2n-5l
     =0
Since, a+b+c=0
Therefore, a³+b³+c³=3abc
Now,
(5l-2m)³+(m-2n)³+(m+2n-5l)³
= a³+b³+c³
= 3abc
= 3(5l-2m)(m-2n)(m+2n-5l)

(xx) 9x²y+90zx²+10y²z+3xy²+300z²x+100yz²+90xyz
Solution:
9x²y+90zx²+10y²z+3xy²+300z²x+100yz²+90xyz
= 9x²y+3xy²+10y²z+90zx²+100yz²+300z²x+90xyz
= 3xy(3x+y)+10yz(y+10z)+30zx(10z+3x)+90xyz
= 3x.y(3x+y)+y.10z(y+10z)+10z.3x(10z+3x)+3.3x.y.10z
= (3x+y+10z)(3x.y+y.10z+10z.3x)
= (3x+y+10z)(3xy+10yz+30zx)

(xxi) ab(a+b-x)+bc(b+c-x)+ca(c+a-x)+3abc
Solution:
ab(a+b-x)+bc(b+c-x)+ca(c+a-x)+3abc
= ab(a+b)-abx+bc(b+c)-bcx+ca(c+a)-cax+3abc
= ab(a+b)+bc(b+c)+ca(c+a)+3abc-abx-bcx-cax
= (a+b+c)(ab+bc+ca)-x(ab+bc+ca)
= (ab+bc+ca)(a+b+c-x)

5.

(i) If a+2b=4c then show that a³+8b³-64c³+24abc=0
Solution:
Given, a+2b=4c 
To prove: a³+8b³-64c³+24abc=0
LHS= a³+8b³-64c³+24abc
   = a³+(2b)³+(-4c)³-3.a.2b.(-4c)
   = (a+3b-4c){a²+(2b²)+(-4c)²-a.2b+2b.4c+4c.a}
   = (4c-4c)(a²+4b²+16c²-2ab+8bc+4ca) [given that a+2b=4c]
   = 0(a²+4b²+16c²-2ab+8bc+4ca)
   = 0
   = RHS
Hence, proved.


(ii) If 3s=2(a+b+c) then show that (s-b-c)³+(s-c-a)³+(s-a-b)³+3(b+c-s)(c+a-s)(a+b-s)=0
Solution:
Given, 3s=2(a+b+c)
Let, s-b-c=x
     s-c-a=y
     s-a-b=z
Now, 
x+y+z=s-b-c+s-c-a+s-a-b
     =3s-2a-2b-2c
     =3s-2(a+b+c)
     =3s-3s [given that 3s=2(a+b+c)]
     =0
Since x+y+z=0, then x³+y³+z³=3xyz
Therefore, 
x³+y³+z³=3xyz
=> (s-b-c)³+(s-c-a)³+(s-a-b)³=3(s-b-c)(s-c-a)(s-a-b)
=> (s-b-c)³+(s-c-a)³+(s-a-b)³=-3(b+c-s)(c+a-s)(a+b-s)
=> (s-b-c)³+(s-c-a)³+(s-a-b)³+3(b+c-s)(c+a-s)(a+b-s)=0
Hence, proved.


(iii) If 2x-y=5, then show that 8x³-y³-30xy+27=152
Solution:
Given, 
   2x-y=5
=> (2x-y)³=5³
=> (2x)³-y³-3.2x.y.(2x-y)=5³
=> (2x)³-y³-6xy.5=125 [given that 2x-y=5]
=> (2x)³-y³-30xy=152-27
=> 8x³-y³-30xy+27=152
Hence, proved.       


(iv) If a=y+z, b=z+x, c=x+y then show that a²+b²+c²-bc-ca-ab=x²+y²+z²-yz-zx-xy
Solution:
Given, a=y+z, b=z+x, c=x+y
a+b+c=y+z+z+x+x+y
     =2x+2y+2z
     =2(x+y+z)
a-b=y+z-z-x
   =y-x
b-c=z+x-x-y
   =z-y
c-a=x+y-y-z
   =x-z
To prove: a²+b²+c²-bc-ca-ab=x²+y²+z²-yz-zx-xy
LHS= a²+b²+c²-bc-ca-ab
   = 1/2*2(a²+b²+c²-bc-ca-ab)
   = 1/2(2a²+2b²+2c²-2bc-2ca-2ab)
   = 1/2(a²-2ab+b²+b²-2bc+c²+c²-2ca+a²)
   = 1/2{(a-b)²+(b-c)²+(c-a)²}
   = 1/2{(y-x)²+(z-y)²+(x-z)²}
   = 1/2(y²-2xy+x²+z²-2yz+y²+x²-2zx+z²)
   = 1/2(2x²+2y²+2z²-2xy-2yz-2zx)
   = 1/2*2(x²+y²+z²-xy-yz-zx)
   = x²+y²+z²-xy-yz-zx
   = RHS
Hence, proved. 


(v)If a+b+c=p then prove that (p-3a)³+(p-3b)³+(p-3c)³-3(p-3a)(p-3b)(p-3c)=0
Solution:
Given, a+b+c=p
Let, p-3a=x
     p-3b=y
     p-3c=z
Now,
x+y+z=p-3a+p-3b+p-3c
     =3p-3(a+b+c)
     =3p-3p [given that a+b+c=p]
     =0
Since x+y+z=0 then x³+y³+z³=3xyz
Therefore,
x³+y³+z³=3xyz
=> (p-3a)³+(p-3b)³+(p-3c)³=3(p-3a)(p-3b)(p-3c)
=> (p-3a)³+(p-3b)³+(p-3c)³-3(p-3a)(p-3b)(p-3c)=0
Hence, proved.


(vi) If p=x²-y², q=2xy and r=x²+y² then show that p⁶+q⁶-r⁶+3p²q²r²=0 .
Solution:
Given, p=x²-y²
       => p²=(x²-y²)²
       => p²=x⁴-2x²y²+y⁴

       q=2xy 
       => q²=(2xy)² 
       => q²=4x²y²

       r=x²+y²
       => r²=(x²+y²)²
       => r²=x⁴+2x²y²+y⁴
Now,
p²+q²-r²=x⁴-2x²y²+y⁴+4x²y²-x⁴-2x²y²-y⁴
=0
Therefore,
(p²)³+(q²)³-(r²)³=3.p².q².(-r²)
=> p⁶+q⁶-r⁶=-3p²q²r²
=>p⁶+q⁶-r⁶+3p²q²r²=0
Hence, proved.


(vii) x²{x⁴-(y²-z²)²}+y²{y⁴-(z²-x²)²}+z²{z⁴-(x²-y²)²}+(x²+y²-z²)(y²+z²-x²)(z²+x²-y²)=4x²y²z² prove this.
Solution:
Let, x²+y²-z²=a
     y²+z²-x²=b
     z²+x²-y²=c
Now,
a+b+c=x²+y²-z²+y²+z²-x²+z²+x²-y²
     =x²+y²+z²
a+b=x²+y²-z²+y²+z²-x²
   =2y²
b+c=y²+z²-x²+z²+x²-y²
   =2z²
c+a=z²+x²-y²+x²+y²-z²
   =2x²

To prove: x²{x⁴-(y²-z²)²}+y²{y⁴-(z²-x²)²}+z²{z⁴-(x²-y²)²}+(x²+y²-z²)(y²+z²-x²)(z²+x²-y²)=4x²y²z²
LHS= x²{x⁴-(y²-z²)²}+y²{y⁴-(z²-x²)²}+z²{z⁴-(x²-y²)²}+(x²+y²-z²)(y²+z²-x²)(z²+x²-y²)
   = x²{x⁴-(y²-z²)²}+y²{y⁴-(z²-x²)²}+z²{z⁴-(x²-y²)²}+abc
   = x²{(x²+y²-z²)(x²-y²+z²)}+y²{(y²+z²-x²)(y²-z²+x²)}+z²{(z²+x²-y²)(z²-x²+y²)}+abc
   = x²ac+y²ab+z²bc+abc
   = 1/2*2(x²ac+y²ab+z²bc+abc)
   = 1/2(2x²ac+2y²ab+2z²bc+2abc)
   = 1/2{(c+a)ca+(a+b)ab+(b+c)bc+2abc}
   = 1/2{(c+a)(a+b)(b+c)}
   = 1/2(a+b)(b+c)(c+a)
   = 1/2.2y².2z².2x²
   = 4y²z²x²
   = 4x²y²z²
   = RHS
Hence, proved.


(viii) If x+y+z=3m/4 then prove that 3(m-4x)(m-4y)(m-4z)=(m-4x)³+(m-4y)³+(m-4z)³
Solution:
Given, x+y+z=3m/4 
Let, m-4x=a, m-4y=b ,m-4z=c
Now,
a+b+c=m-4x+m-4y+m-4z
     =3m-4(x+y+z)
     =3m-4*3m/4 [given that x+y+z=3m/4]
     =3m-3m
     =0
Since a+b+c=0 then a³+b³+c³=3abc
Therefore,
a³+b³+c³=3abc
=> (m-4x)³+(m-4y)³+(m-4z)³=3(m-4x)(m-4y)(m-4z)
=> 3(m-4x)(m-4y)(m-4z)=(m-4x)³+(m-4y)³+(m-4z)³
Hence, proved.


(ix) If p=x+y-z, q=y+z-x, r=z+x-y then prove that p³+q³+r³-4(x³+y³+z³)=3(pqr-4xyz)
Solution:
Given, p=x+y-z, q=y+z-x, r=z+x-y
p+q+r=x+y-z+y+z-x+z+x-y
     =x+y+z
p-q=x+y-z-y-z+x
   =2(x-z)
q-r=y+z-x-z-x+y
   =2(y-x)
r-p=z+x-y-x-y+z
   =2(z-y)
We know,
   p³+q³+r³-3pqr = 1/2(p+q+r)[(p-q)²+(q-r)²+(r-p)²]
                 = 1/2(x+y+z)[{2(x-z)}²+{2(y-x)}²+{2(z-y)}²]
                 = 1/2(x+y+z)[4(x-z)²+4(y-x)²+4(z-y)²]
                 = 1/2(x+y+z)*4[(x-z)²+(y-x)²+(z-y)²]
                 = 4*1/2(x+y+z)[(x-z)²+(y-x)²+(z-y)²]
                 = 4(x³+y³+z³-3xyz)
=> p³+q³+r³-3pqr = 4(x³+y³+z³)-12xyz
=> p³+q³+r³-4(x³+y³+z³)=3pqr-12xyz
=> p³+q³+r³-4(x³+y³+z³)=3(pqr-4xyz)
Hence, proved.












































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