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+2a^2b^2+2b^2c^2+2c^2a^2

3 Answers
May 31, 2018

a^4+ab^3=a(a+b)(a^2-ab+b^2)

Explanation:

Multiplying

(a+b)(a^2-ab+b^2)
out we get

a^3+a^2b-a^2b-ab^2+ab^2+b^3
For 2)

16s(s-a)(s-b)(s-c)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2
(s=(a+b+c)/2)

May 31, 2018

rarra^4+ab^3=a(a^3+b^3)=a*(a+b)(a^2-ab+b^2)

rarra^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2

=(a^2)^2+2a^2*b^2+(b^2)^2+2c^2(a^2+b^2)

=(a^2+b^2)^2+2c^2(a^2+b^2)

=(a^2+b^2)(a^2+b^2+2c^2)

May 31, 2018

1) a(a+b)(a^2-ab+b^2)
2) Unfactorisable

Explanation:

1) a^4+ab^3

Using exponent rule: a^(b+c)=a^b*a^c

a^4=aa^3
aa^3+ab^3

Factor out common term, a

a(a^3+ b^3)

Apply sum of two cubes rule: x^3+y^3=(x+y)(x^2-xy+y^2)

a^3+b^3=(a+b)(a^2-ab+b^2)

therefore a^4+ab^3=a(a+b)(a^2-ab+b^2)

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