Class 9 Advance Maths(Elective) Chapter 4 Exercise 4.2 (part 2)

 

4. SPECIAL PRODUCT AND FACTORIZTION

Exercise 4.2

For the previous solutions Click here

6. 

(i) If x+y+z=6, x²+y²+z²=14 then what will be the value of x³+y³+z³-3xyz?

Solution:
Given,  x+y+z=6,  x²+y²+z²=14
We know,
(x+y+z)²=x²+y²+z²+2xy+2yz+2zx
=> 6²=14+2(xy+yz+zx)
=> 36-14=2(xy+yz+zx)
=> 2(xy+yz+zx)=22
=> xy+yz+zx=22/2
=> xy+yz+zx=11
Therefore,
  x³+y³+z³-3xyz
= (x+y+z)(x²+y²+z²-xy-yz-zx)
= 6{14-(xy+yz+zx)}
= 6(14-11)
= 6*3
= 18


(ii) If x=p+1, y=p+2, z=p+3 then find yhe value x³+y³+z³-3xyz.

Solution:
Given,  x=p+1, y=p+2, z=p+3
x+y+z=p+1+p+2+p+3
     =3p+6
     =3(p+2)
x-y=p+1-p-2
   =-1
y-z=p+2-p-3
   =-1
z-x=p+3-p-1
   =2
Therefore,
  x³+y³+z³-3xyz
= 1/2(x+y+z){(x-y)²+(y-z)²+(z-x)²}
= 1/2*3(p+2){(-1)²+(-1)²+2²}
= 1/2*3(p+2)(1+1+4)
= 1/2*3(p+2)*6
= 18/2(p+2)
= 9(p+2)


(iii) a+b+c=10, bc+ca+ab=15 and abc=25 then find the value of a²(b+c)+b²(c+a)+c²(a+b).

Solution:
Given,  a+b+c=10, bc+ca+ab=15 and abc=25
Therefore,
  a²(b+c)+b²(c+a)+c²(a+b)
= (a+b+c)(bc+ca+ab)-3abc
= 10*15-3*25
= 150-75
= 75


(iv) a+b+c=15, a²+b²+c²=83 and a³+b³+c³=495 then find the value of abc. Also ab+bc+ca.

Solution:
Given, a+b+c=15, a²+b²+c²=83 and a³+b³+c³=495
We know that,
   (a+b+c)²=a²+b²+c²+2ab+2bc+2ca
=> 15²=83+2(ab+bc+ca)
=> 225-83=2(ab+bc+ca)
=> 142=2(ab+bc+ca)
=> 142/2=ab+bc+ca
=> ab+bc+ca=142/2
=> ab+bc+ca=71
Again,
   a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
=> 495-3abc=15(83-71)
=> 495-3abc=15*12
=> 495-3abc=180
=> 495-180=3abc
=> 3abc=315
=> abc=315/3
=> abc=105


(v) If a+b+c=12, a²+b²+c²=6, a³+b³+c³=9 then find the value of abc and ab+bc+ca.

Solution:
Given, a+b+c=12, a²+b²+c²=6, a³+b³+c³=9
We know that,
   (a+b+c)²=a²+b²+c²+2ab+2bc+2ca
=> 12²=6+2(ab+bc+ca)
=> 144-6=2(ab+bc+ca)
=> 138/2=ab+bc+ca
=> ab+bc+ca=69
Again,
   a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
=> 9-3abc=12{6-(ab+bc+ca)}
=> 9-3abc=12(6-69)
=> 9-3abc=12*(-63)
=> 9-3abc=-756
=> 9+756=3abc
=> 3abc=765
=> abc=765/3
=> abc=255


7. If x²-yz , y²-zx , z²-xy are mutually equal then show tht the product of (p+q+r) and (x+y+z) is
          p        q          r 
    equal to px+qy+rz . 
       

Solution: We have,
x²-yz = y²-zx = z²-xy
     p           q         r
Let,
   x²-yz = y²-zx = z²-xy = k
      p           q         r

=> x²-yz = k  ,              => y²-zx = k ,             => z²-xy = k
        p                                  q                                r

=> x²-yz = kp ------(i) , =>y²-zx = kq------(ii) , =>z²-xy = kr------(iii)

Therefore,
(i)*x+(ii)*y+(iii)*z
=> (x²-yz)x+(y²-zx)y+(z²-xy)z = kpx+kqy+krz
=> x³-xyz+y³-xyz+z³-xyz = k(px+qy+rz)
=> x³+y³+z³-3xyz = k(px+qy+rz)
=> (x+y+z)(x²+y²+z²-xy-yz-zx) = k(px+qy+rz)
=> (x+y+z)(x²-yz+y²-zx+z²-xy) = k(px+qy+rz)
=> (x+y+z)(kp+kq+kr) = k(px+qy+rz)
=> (x+y+z)*k(p+q+r) = k(px+qy+rz)
=> (x+y+z)(p+q+r) = px+qy+rz
Hence, proved.

8. If the equation (x+a)(x+b)(x+c)=0 is same with the equation x³+5x²+8x-11=0 then find the value of a²(b+c)+b²(c+a)+c²(a+b).
Solution: 
Given,
   (x+a)(x+b)(x+c)=0
=> x³+x²(a+b+c)+x(ab+bc+ca)+abc=0------(i)
Since, 
x³+5x²+8x-11=0
Therefore, equation (i)=x³+5x²+8x-11
From the above prove, 
We get, a+b+c=5 , ab+bc+ca=8 ,and abc=-11
Now,
  a²(b+c)+b²(c+a)+c²(a+b)
= (a+b+c)(ab+bc+ca)-3abc
= 5*8-3*11
= 40-33
= 7
Hence, the value of a²(b+c)+b²(c+a)+c²(a+b) is 7.


9. If the equation (x+a)(x+b)(x+c)=0 is same with the equation x³-9x²+5x-13=0 then find the value of bc(b+c)+ca(c+a)+cab(a+b).
Solution:
Given,  (x+a)(x+b)(x+c)=0
     => x³+x²(a+b+c)+x(ab+bc+ca)+abc=0------(i)
Since, 
x³-9x²+5x-13=0
Therefore, equation (i)=x³-9x²+5x-13
From the above prove, 
We get, a+b+c=-9 , ab+bc+ca=5 , abc=-13
Now,
  bc(b+c)+ca(c+a)+cab(a+b)
= (a+b+c)(ab+bc+ca)-3abc
= -9*5+13*3
= -45+39
= -6


10. If a+b+c=2p then show that 8p³=a³+b³+c³+3(2p-a)(2p-b)(2p-c).
Solution:
Given,
  a+b+c=2p
=> a+b=2p-c , => b+c=2p-a , => c+a=2p-b

We know that,
   a+b+c=2p
=> (a+b+c)³=(2p)³
=> a³+b³+c³+3(a+b)(b+c)(c+a)=8p³
=> a³+b³+c³+3(2p-c)(2p-a)(2p-b)=8p³
=> 8p³=a³+b³+c³+3(2p-a)(2p-b)(2p-c)
Hence, proved.


11. Simplify -

(i) 2a(b+c-a)²+2b(c+a-b)²+2c(a+b-c)²+2(a+b+c) [{a²-(b-c)²}+{b²-(c-a)²}+{c²-(a-b)²}]
Solution: 
We know,
  2a(b+c-a)²+2b(c+a-b)²+2c(a+b-c)²+2(a+b+c) [{a²-(b-c)²}+{b²-(c-a)²}+{c²-(a-b)²}]
= 2a(b+c-a)²+2b(c+a-b)²+2c(a+b-c)²+2(a+b+c)[(a-b+c)(a+b-c)+(b-c+a)(b+c-a)+(c-a+b)(c+a-b)]
= 2a(b+c-a)²+2b(c+a-b)²+2c(a+b-c)²+2(a+b+c)[(c+a-b)(a+b-c)+(a+b-c)(b+c-a)+(b+c-a)(c+a-b)]
Let, (b+c-a)=x
     (c+a-b)=y
     (a+b-c)=z
x+y+z=(b+c-a)+(c+a-b)+(a+b-c)
     =b+c-a+c+a-b+a+b-c
     =a+b+c
x+y=b+c-a+c+a-b
   =2c
y+z=c+a-b+a+b-c
   =2a
z+x=a+b-c+b+c-a
   =2b
Therefore,the given expression is-
  2a(b+c-a)²+2b(c+a-b)²+2c(a+b-c)²+2(a+b+c)[(c+a-b)(a+b-c)+(a+b-c)(b+c-a)+(b+c-a)(c+a-b)]
= (y+z)x²+(z+x)y²+(x+y)z²+2(x+y+z)(y.z+z.x+x.y)
= (y+z)x²+(z+x)y²+(x+y)z²+2(x+y+z)(xy+yz+zx)
= (y+z)x²+(z+x)y²+(x+y)z²+2(x+y+z)(xy+yz+zx)+3xyz-3xyz
= (y+z)x²+(z+x)y²+(x+y)z²+3xyz+2(x+y+z)(xy+yz+zx)-3xyz
= (x+y+z)(xy+yz+zx)+2(x+y+z)(xy+yz+zx)-3xyz
= 3(x+y+z)(xy+yz+zx)-3xyz
= 3{(x+y+z)(xy+yz+zx)-xyz}
= 3(x+y)(y+z)(z+x)
= 3.2a.2b.2c
= 24abc


(ii) x(x-y+z)(x+y-z) + y(y-z+x)(y+z-x) + z(z-x+y)(z+x-y) + (x+y-z)(y+z-x)(z+x-y)
Solution:
We know,
  x(x-y+z)(x+y-z) + y(y-z+x)(y+z-x) + z(z-x+y)(z+x-y) + (x+y-z)(y+z-x)(z+x-y)
= x(z+x-y)(x+y-z) + y(x+y-z)(y+z-x) + z(y+z-x)(z+x-y) + (x+y-z)(y+z-x)(z+x-y)
Let,
x+y-z=a
y+z-x=b
z+x-y=c
a+b+c=x+y-z+y+z-x+z+x-y
     =x+y+z
a+b=x+y-z+y+z-x
   =2y
b+c=y+z-x+z+x-y
   =2z
c+a=z+x-y+x+y-z
   =2x
Therefore, the given expression is-
  x(z+x-y)(x+y-z) + y(x+y-z)(y+z-x) + z(y+z-x)(z+x-y) + (x+y-z)(y+z-x)(z+x-y)
= xac+yba+zcb+abc
= 1/2*2(xca+yab+zbc+abc)
= 1/2(2xca+2yab+2zbc+2abc)
= 1/2[ab(a+b)+bc(b+c)+ca(c+a)+2abc]
= 1/2(a+b)(b+c)(c+a)
= 1/2*2y*2z*2x
= 4xyz


(iii) (x-y)²(y-z) + (y-z)²(z-y) + (z-x)²(x-z) + 3(x-y)(y-z)(z-x)
Solution:
Let, x-y=a
     y-z=b
     z-x=c
a+b+c=x-y+y-z+z-x
     =0
a+b=x-y+y-z
   =x-z
b+c=y-z+z-x
   =y-z
c+a=z-x+x-y
   =z-y
Therefore, the given expression is-
  (x-y)²(y-z) + (y-z)²(z-y) + (z-x)²(x-z) + 3(x-y)(y-z)(z-x)
= a²(b+c)+b²(c+a)+c²(a+b)+3abc
= (a+b+c)(ab+bc+ca)
= 0*(ab+bc+ca)
= 0












Comments

Post a Comment

If you have any doubts, please let me know

Popular posts from this blog

Class 10 Geography Chapter 1

CLASS 10 SEBA SOCIAL STUDIES NOTES ECONOMIC GEOGRAPHY : SUBJECT MATTER AND RESOURCE 1.What is economic geography? What is its main subject matter? Mention the important branches of Economic Geography. Ans:-  The branch of Geography in which the study of human activities associated with production, distribution, consumption and exchange of resources is done in spatial and temporal context is known as Economic Geography. The main subject matter of economic geography is to study the extent of diversity and prevalence of the variation in the elements such as natural and manmade environment.  The important branches of Economic Geography are - agricultural geography, industrial geography, resource geography, transport geography, geography of marketing, geography of planning and development, and geography of tourism. 2. Write in brief about the scope of Economic Geography. Ans:-  The distribution of economic activities of man and the factors and processes associated with it are ...
Read more

Class 10 History Chapter 1

     CLASS 10 SEBA SOCIAL STUDIES NOTES 1. Partition of Bengal and Swadeshi Movement Very short answer questions: 1. In which Viceroy's time did the partition of Bengal take place? Ans:-  The partition of Bengal took place in Lord Curzon's time. 2. On which date in 1905 did the partition of Bengal come into force?  Ans:-   On 16th October in 1905 the partition of Bengal came into force. 3. In which meeting was the proposal of giving up foreign commodities accepted?  Ans:-   In the 2nd protest meeting held at Dinazpur on 20th July, 1905 the proposal of giving up foreign commodities was accepted. 4. Who was the editor of the journal Jugaantar published in the time of Swadeshi movement?  Ans:-  Bhupendra Nath Dutta was the editor of the journal Jugaantar published in the time of Swadeshi movement. 5. When was the institution named National Education Council (Jatiya Siksha Parisad) established in Bengal?  Ans:-  The institution n...
Read more

Class 10 Geography Chapter 2 Seba Notes |

GEOGRAPHY CHAPTER-2 ENVIRONMENT AND ENVIRONMENTAL PROBLEMS 1. Write in your own words the meaning of environment. Ans: The condition created by the interaction among the four spheres, ie lithosphere , hydrosphere , atmosphere and biosphere may broadly be called environment. It denotes the condition around an organism or a community of organism and such a condition includes all the necessary elements for survival. These elements are both biotic and abiotic. These biotic and abiotic elements through their interaction create necessary condition or environment for life. 2. Mention some of the biotic and abietic elements of environment.  Ans: Biotic- Plante, animals, ranging from microbes to large mammals. Abiotic - Land, water, air including all lifeless objects.  3. Write the meaning of lithosphere.  Ans: The Land component of the earth composed of rocks , soils, minerals etc forms the Lithosphere. The lithosphere covers all the contents where we find the mountains chills, ...
Read more