Question

4(a2-b2)2-8ab(a2-b2)-5a2b2 please help me​

Answer 1

Answer: $4a^4+4b^4-8a^3b+8ab^3-13a^2b^2$

Step-by-step explanation:

$4(a^2-b^2)^2-8ab(a^2-b^2)-5a^2b^2$

First, solve the square binomial.

$4((a^2)^2-2(a^2)(b^2)+(b^2)^2)-8ab(a^2-b^2)-5a^2b^2$

$4(a^4-2a^2b^2+b^4)-8ab(a^2-b^2)-5a^2b^2$

Now, distribute the 4 and 8ab respectively.

$4a^4-8a^2b^2+4b^4-8a^3b+8ab^3-5a^2b^2$

Combine like terms. (I'll group them so that you can see them more clearly)

$4a^4+4b^4-8a^3b+8ab^3+(-8a^2b^2-5a^2b^2)$

$4a^4+4b^4-8a^3b+8ab^3+(-13a^2b^2)$

$4a^4+4b^4-8a^3b+8ab^3-13a^2b^2$

Answer 2

Answer:

(2a²-2b²+a²b²)(2a²-2b²-5a²b²)

Step-by-step explanation:

4(a²-b²)²-8ab(a²-b²)-5a²b²

Let's replace 2(a²-b²)= m and a²b²=n for simplicity

then we have

• m²-4mn-5n²
• = m²-4mn+4n²-4n²-5n²
• = (m-2n)²- 9n²
• = (m-2n)²- (3n)²
• = (m-2n+3n)(m-2n-3n)
• = (m+n)(m-5n)

Now we can replace m and n with initial values:

• (m+n)(m-5n)=
• = (2(a²-b²)+a²b²)(2(a²-b²)- 5a²b²)
• = (2a²-2b²+a²b²)(2a²-2b²-5a²b²)