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
Sonnhard
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#)

Abhishek K.
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)#

Layomi.O
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.

Related questions
Impact of this question
2604 views around the world
You can reuse this answer
Creative Commons License