Chapter B, Problem 1E

To determine

To find:

The factorisation of the given binomial.

Answer to Problem 1E

Solution:

The given binomial $a^3+27$ can be factorized as $\left(a+3\right)\left(a^2-3a+9\right)$.

Explanation of Solution

Formula Used:

1) Factor out common factor from both the terms.

2) By the formula of sum of two cubes,

$u^3+v^3=\left(u+v\right)\left(u^2-uv+v^2\right)$

3) By the formula of difference of two cubes,

$u^3-v^3=\left(u-v\right)\left(u^2+uv+v^2\right)$

Calculation:

The given binomial is $a^3+27$.

Use the formula of sum of two cubes,

$u^3+v^3=\left(u+v\right)\left(u^2-uv+v^2\right)$

Substitute u = a and v = 3 in $u^3+v^3=\left(u+v\right)\left(u^2-uv+v^2\right)$

$\begin{array}{l}{a}^3+3^3=\left(a+3\right)\left(a^2-a\cdot 3+3^2\right)\\ \Rightarrow a^3+27=\left(a+3\right)\left(a^2-3a+9\right)\end{array}$

Therefore, the given binomial $a^3+27$ can be factorized as $\left(a+3\right)\left(a^2-3a+9\right)$.

Final statement:

The given binomial $a^3+27$ can be factorized as $\left(a+3\right)\left(a^2-3a+9\right)$.