NCERT Class 9 Solutions: Polynomials (Chapter 2) Exercise 2.5 – Part 3

Questioning the algebra

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Questioning the algebra

Q-9 Verify:

  1. x3+y3=(x+y)(x2xy+y2)

  2. x3y3=(xy)(x2+xy+y2)

Solution:

  1. x3+y3=(x+y)(x2xy+y2)

    R.H.S =x(x2xy+y2)+y(x2xy+y2)

    =x3+x2y+xy2+yx2xy2+y3

    =x3+y3

    L.H.S

  2. x3y3=(xy)(x2+xy+y2)

    R.H.S =x(x2+xy+y2)y(x2+xy+y2)

    =x3+x2y+xy2yx2xy2y3

    =x3y3

    L.H.S

Q-10 Factorize each of the following:

  1. 27y3+125z3

  2. 64m3343n3

Solution:

  1. 27y3+125z3

    =(3y)3+(5z)3

    Using identify x3+y3=(x+y)(x2+y2xy)

    =(3y+5z)((3y)2+(5z)2(3y)(5z))

    =(3y+5z)(9y2+25z215yz)

  2. 64m3343n3

    =(4m)3(7n)3

    Using identify x3+y3=(x+y)(x2+y2xy)

    =(4m7n)((4m)2+(7n)2(4m)(7n))

    =(4m7n)(16m2+49n2+28mn)

Q-11 Factorize: 27x3+y3+z39xyz

Solution:

Using x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)

(3x)3+y3+z33(3x)yz

=(3x+y+z)((3x)2+y2+z23xyyz3zx)

=(3x+y+z)(9x2+y2+z23xyyz3zx)

Q-12 Verify that:

x3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2]

Solution:

RHS=12(x+y+z)[x2+y22xy+y2+z22yz+z2+x22zx]

=12(x+y+z)[2x2+2y2+2z22xy2yz2zx]

=(x+y+z)[x2+y2+z2xyyzzx]

=x[x2+y2+z2xyyzzx]+y[x2+y2+z2xyyzzx]+z[x2+y2+z2xyyzzx]

=x3+xy2+xz2x2yxyzzx2+x2y+y3+yz2xy2y2zxyz+x2z+y2z+z3xyzyz2z2x

=x3+y3+z33xyz

=L.H.S

Q-13 If x+y+z=0 , show that x3+y3+z3=3xyz

Solution:

x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)

Also it is given x+y+z=0

x3+y3+z33xyz=(0)(x2+y2+z2xyyzzx)

x3+y3+z33xyz=0

x3+y3+z3=3xyz

Q-14 Without actually calculating the cubes fined the value of each of the following:

  1. (12)3+(7)3+(5)3

  2. (28)x3+(15)x3+(13)x3

Solution:

  1. (12)3+(7)3+(5)3

    (12)+(7)+(5)=0

    Using identify, if x+y+z=0,thenx3+y3+z3=3xyz

    =3(12)(7)(5)

    =1260

  2. (28)x3+(15)x3+(13)x3

    (28)+(15)+(13)=0

    Using identify, if x+y+z=0,thenx3+y3+z3=3xyz

    =3(28)(15)(13)

    =16380

Q-15 Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

  1. Area:25a235a+12

  2. Area:35y2+13y12

Solution:

Since areaofrectangle=length×breadth

Let us factorize the following equations into two terms

  1. Area:25a235a+12

    =25a215a20a+12

    =5a(5a3)4(5a3)

    =(5a3)(5a4)

    Possiblelength=(5a3)

    andpossiblewidth=(5a4)

  2. Area:35y2+13y12

    =35y2+28y15y12

    =7y(5y+4)3(5y+4)

    =(5y+4)(7y3)

    Possiblelength=(5y+4)

    andpossiblewidth=(7y3)

Q-16 What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

  1. Volume:3x212x

  2. Volume:12ky2+8ky20k

Solution:

Since areaofcuboid=length×breadth×height

Let us factorize the following equations into two terms

  1. Volume:3x212x

    =3x(x4)

    Possiblelength=3,width=xandheight=(x4)

    orPossiblelength=1,width=3xandheight=(x4)

  2. Volume:12ky2+8ky20k

    =4k(3y2+2y5)

    =4k(3y23y+5y5)

    =4k(3y(y1)+5(y1))

    =4k(y1)(3y+5)

    Possiblelength=4k,width=(y1)andheight=(3y+5)

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