Question

(2a-3b)³+64c³ factorise ​

Answer 1

Answer:

You can do that by using identity

${(a - b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$

Then you will get your answer

Answer 2

$\huge\mathbb{BONJOUR!!}$

$\huge\bold\pink{Answer:-}$

⏩(2a-3b)³ + 64c³

Using a³+b³= (a+b)(a²-ab+b²), factor out the expression...

→(2a-3b+4c)×[(2a-3b)² -(2a-3b)×4c + 16c²]

Using (a-b)²= a²-2ab+b², expand the expression and then distribute 4c through the bracket for the next expression...

→(2a-3b+4c)×(4a²-12ab+9b²-8ac+12bc+16c²)

The required factorization...

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Hope it helps ..:-)

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