Question

factorise a⁴-(a-b)⁴​

Answer 1

Answer:

We know, (a+b)⁴ = a⁴+4a³b+6a²b²+4ab³+b⁴

Also, (a-b)⁴ = a⁴-4a³b+6a²b²-4ab³+b⁴

So, (a+b)⁴-(a-b)⁴

= a⁴+4a³b+6a²b²+4ab³+b⁴-a⁴+4a³b-6a²b+4ab³-b⁴

= 8 a³b + 8 ab³

= 8ab(a²+b²) [Answer]

Answer 2

The factories form is $a^4-(a-b)^4=(2a^2+b^2-2ab)(2a-b)(-b)$

Step-by-step explanation:

Given : Expression $a^4-(a-b)^4$

To find : Factories the expression ?

Solution :

Re-write the expression as,

$a^4-(a-b)^4=(a^2)^2-((a-b)^2)^2$

Using algebraic identity, $a^2-b^2=(a+b)(a-b)$

$a^4-(a-b)^4=(a^2+(a-b)^2)(a^2-(a-b)^2)$

Again using same identity,

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

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

Therefore, the factories form is $a^4-(a-b)^4=(2a^2+b^2-2ab)(2a-b)(-b)$

#Learn more

(b+c) /(a-b) (a-c) +(c+a) /(b+a) (b-c) +(a+b) /(c-a) (c-b)

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