# Factorise the following by using suitable identities?

## Factorise 1)a^4+ab^3 Factorise 1) ${a}^{4} + a {b}^{3}$ 2)${a}^{4} + {b}^{4} + {c}^{4} + 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2}$

May 31, 2018

${a}^{4} + a {b}^{3} = a \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

#### Explanation:

Multiplying

$\left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
out we get

${a}^{3} + {a}^{2} b - {a}^{2} b - a {b}^{2} + a {b}^{2} + {b}^{3}$
For 2)

$16 s \left(s - a\right) \left(s - b\right) \left(s - c\right) = {a}^{4} + {b}^{4} + {c}^{4} + 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2}$
($s = \frac{a + b + c}{2}$)

May 31, 2018

$\rightarrow {a}^{4} + a {b}^{3} = a \left({a}^{3} + {b}^{3}\right) = a \cdot \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

$\rightarrow {a}^{4} + {b}^{4} + {c}^{4} + 2 {a}^{2} {b}^{2} + 2 {b}^{2} {c}^{2} + 2 {c}^{2} {a}^{2}$

$= {\left({a}^{2}\right)}^{2} + 2 {a}^{2} \cdot {b}^{2} + {\left({b}^{2}\right)}^{2} + 2 {c}^{2} \left({a}^{2} + {b}^{2}\right)$

$= {\left({a}^{2} + {b}^{2}\right)}^{2} + 2 {c}^{2} \left({a}^{2} + {b}^{2}\right)$

$= \left({a}^{2} + {b}^{2}\right) \left({a}^{2} + {b}^{2} + 2 {c}^{2}\right)$

May 31, 2018

1) $a \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
2) Unfactorisable

#### Explanation:

1) ${a}^{4} + a {b}^{3}$

Using exponent rule: ${a}^{b + c} = {a}^{b} \cdot {a}^{c}$

${a}^{4}$=$a {a}^{3}$
$a {a}^{3} + a {b}^{3}$

Factor out common term, $a$

$a \left({a}^{3} + {b}^{3}\right)$

Apply sum of two cubes rule: ${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

$\therefore {a}^{4} + a {b}^{3} = a \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

2) Unfortunately that question is already factorised to its lowest.