Question

Factorize the following.
27x³+64x³

Answer 1

Hope this helps you out.

Answer 2

We have to use the identity given below.

$\Large a^3+b^3=(a+b)(a^2-ab+b^2)$

So,

$27x^3+64x^3\\ \\ \\ (3x)^3+(4x)^3\\ \\ \\ (3x+4x)((3x)^2-3x \cdot 4x+(4x)^2)\\ \\ \\ 7x(9x^2-12x^2+16x^2) \\ \\ \\ 7x \cdot 13x^2$

But we only get these two factors by this method.

We can also do the following:

It is seen that,

$27x^3+64x^3=91x^3$

91 can be split as  1 × 91  and  7 × 13.

And x³ can be split as  1 · x³  and  x · x².

From these factors, we can factorize 27x³ + 64x³ in '8' ways:

$1.\ (1\cdot 1) \cdot (91 \cdot x^3)=\bold{1 \cdot 91x^3}\\ \\ 2.\ (1 \cdot x^3) \cdot (91 \cdot 1)=\bold{x^3 \cdot 91}\\ \\ 3.\ (1 \cdot x)\cdot (91 \cdot x^2)=\bold{x \cdot 91x^2}\\ \\ 4.\ (1 \cdot x^2)\cdot (91 \cdot x)=\bold{x^2 \cdot 91x}\\ \\ 5.\ (7\cdot 1) \cdot (13 \cdot x^3)=\bold{7 \cdot 13x^3}\\ \\ 6.\ (7 \cdot x^3) \cdot (13 \cdot 1)=\bold{7x^3 \cdot 13}\\ \\ 7.\ (7 \cdot x)\cdot (13 \cdot x^2)=\bold{7x \cdot 13x^2}\ \ \ \ \ [\text{Got earlier}] \\ \\ 8.\ (7 \cdot x^2)\cdot (13 \cdot x)=\bold{7x^2 \cdot 13x}$