Question

$64 {a}^{3} - 27 {b}^{2} - 144 {a}^{2} b + 108a {b}^{2}$


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Answer 1

Answer:

Step-by-step explanation:

64a³ - 27b³ - 144a²b + 108ab²

Here, using the identity -

(a - b)³ = a³ - b³ - 3a²b + 3ab²

⇒ 64a³ - 27b³ - 144a²b + 108ab² = (4a)³ - (3b)³ - 3(4a)²*(3b) + 3(4a)*(3b)²

⇒ (4a - 3b)³

⇒ (4a - 3b)*(4a - 3b)*(4a - 3b)

Answer 2

$\underline{\mathfrak{\huge{Question:}}}$

Factorise the equation :-

$64a^{3} - 27b^{3} - 144a^{2} b + 108ab^{2}$

$\underline{\mathfrak{\huge{Answer:}}}$

Let's start with simplifying the given equation :-

$\tt{(4a)^{3} - (3b)^{3} - (3)(4a)^{2}(3b) + (3)(4a)(3b)^{2}}$

Now, by seeing and observing this new formed equation, we can say that the identity :-

$\tt{(a-b)^{3} = a^{3} - b^{3} - 3a^{2}b + 3ab^{2}}$

Has been used here

Where, a will be = 4a

And, b will be = 3b

So, by all this, we can easily write that :-

$64a^{3} - 27b^{3} - 144a^{2} b + 108ab^{2}$ = $\tt{(4a)^{3} - (3b)^{3} - (3)(4a)^{2}(3b) + (3)(4a)(3b)^{2}}$ = $\tt{(4a-3b)^{3}}$

There's your answer. Its the final result.