Question

Factorize: (i) x3 + 216y3 + 8z3 – 36xyz

(ii) a3 – 64b3 – 27c3 – 36abc

Answer 1

Given: x^3 + 216y^3 + 8z^3 – 36xyz and a^3 – 64b^3 – 27c^3 – 36abc

To find: Factorise the given terms.

Solution:

• Now we know that:

               a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc -ac)

• The first term is:

             x^3 + 216y^3 + 8z^3 – 36xyz

• It can be written as:

               x^3 + 6y^3 + 2z^3 – 3(x)(6y)(2z)  

               (x + 6y + 2z) { (x)^2 + (6y)^2 + (2z)^2 - (x)(6y) - (6y)(2z) - (x)(2z)] }         

               (x + 6y + 2z) (x^2 + 36y^2 + 4z^2 - 6xy - 12yz - 2xz)

• The second term is:

               a^3 – 64b^3 – 27c^3 – 36abc

• It can be written as:

               a^3 + (-4b)^3 + (-3c)^3 - 3(a)(-4b)(-3c)

               (a - 4b - 3c) { (a)^2 + (-4b)^2 + (-3c)^2 - (a)(-4b) - (-4b)(-3c) - (a)(-3c) }  

               (a - 4b - 3c) { a^2 + 16b^2 + 9c^2 + 4ab - 12bc + 3ac }  

Answer:

           So the factorisation is done in solution part.

Answer 2

GIVEN :

Factorize the expressions

(i) $x^3 + 216y^3 + 8z^3-36xyz$

(ii)  $a^3 -64b^3 - 27c^3 -36abc$

TO FIND :

The simplified factors for the given expressions

SOLUTION :

Given expressions are

(i) $x^3 + 216y^3 + 8z^3-36xyz$

(ii)  $a^3 -64b^3 - 27c^3 -36abc$

Now factorize (i) $x^3 + 216y^3 + 8z^3-36xyz$

$=x^3 + 6^3y^3 + 2^3z^3-36xyz$

By using the exponent property :

$(ab)^m=a^mb^m$

$=x^3 + (6y)^3 + (2z)^3-36xyz$

$=x^3 + (6y)^3 + (2z)^3-3(x)(6y)(2z)$

Here a=x , b=6y and c=2z

By using the Algebraic identity :

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-2ab-2bc-2ca)$

$=(x+6y+2z)(x^2+(6y)^2+(2z)^2-2(x)(6y)-2(6y)(2z)-2(2z)(x))$

By using the exponent property :

$(ab)^m=a^mb^m$

$=(x+6y+2z)(x^2+6^2(y)^2+2^2(z)^2-12xy-24yz-4zx)$

$=(x+6y+2z)(x^2+36y^2+4z^2-12xy-24yz-4zx)$

$=(x+6y+2z)[x(x-12y)+12y(3y-2z)+4z(z-x)]$

∴ $x^3 + 216y^3 + 8z^3-36xyz=(x+6y+2z)[x(x-12y)+12y(3y-2z)+4z(z-x)]$

∴ the given polynomial  $x^3 + 216y^3 + 8z^3-36xyz$ is factorized into the expression $(x+6y+2z)[x(x-12y)+12y(3y-2z)+4z(z-x)]$

Now factorize (ii) $a^3 -64b^3 - 27c^3 -36abc$

$=a^3 -64b^3 - 27c^3 -36abc$

$=a^3 -4^3b^3 - 3^3c^3 -36abc$

By using the exponent property :

$(ab)^m=a^mb^m$

$=a^3 + (-4b)^3 + (-3c)^3-36abc$

$=a^3 + (-4b)^3 + (-3c)^3-3(a)(-4b)(-3c)$

Here x=a , y=-4b and z=-3c

By using the Algebraic identity :

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-2xy-2yz-2zx)$

$=(a-4b-3c)(a^2+(-4b)^2+(-3c)^2-2(a)(-4b)-2(-4b)(-3c)-2(-3c)(a))$

$=(a-4b-3c)(a^2+(4b)^2+(3c)^2-2(a)(-4b)-2(-4b)(-3c)-2(-3c)(a))$

By using the exponent property :

$(ab)^m=a^mb^m$

$=(a-4b-3c)(a^2+4^2b^2+3^2(c)^2+8ab-24bc+6ca)$

$=(a-4b-3c)(a^2+16b^2+9(c)^2+8ab-24bc+6ca)$

$=(a-4b-3c)[a(a+8b)+8b(2b-3c)+3c(3c+2a)]$

∴ $a^3 -64b^3 - 27c^3 -36abc=(a-4b-3c)[a(a+8b)+8b(2b-3c)+3c(3c+2a)]$

∴ the given polynomial  $a^3 -64b^3 - 27c^3 -36abc$ is factorized into the expression $(a-4b-3c)[a(a+8b)+8b(2b-3c)+3c(3c+2a)]$