Question

What is the formula of (a-b) whole cube?

Answer 1

Answer:

(a-b)³ = a³-3a²b+3ab²-b³

Or

(a-b)³ = a³-b³-3ab(a-b)

Explanation:

(a-b)³

= (a-b)(a-b)²

= (a-b)( a²-2ab+b²)

/* we know the algebraic identity:

(a-b)² = a²-2ab+b² */

= a(a²-2ab+b²)-b(a²-2ab+b²)

= a³-2a²b+ab²-a²b+2ab²-b³

= a³ - 3a²b + 3ab² - b³

Or

= a³-b³-3ab(a-b)

Therefore,

(a-b)³ = a³ - 3a²b + 3ab² - b³

Or

(a-b)³ = a³ - b³ - 3ab(a-b)

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Answer 2

The formula of $(a-b)^{3} \text { is } \bold{\left(a^{3}-3 a^{2} b+3 a b^{2}-b^{3}\right)}$.

To find:

Formula of (a-b)^{3}

Solution:

$(a-b)^{3}$

$=(a-b)(a-b)(a-b)$

$=[a(a-b)-b(a-b)](a-b)$

$=\left(a^{2}-a b-b a+b^{2}\right)(a-b)$

We know that ab = ba. Hence,

$\begin{array}{l}{=\left(a^{2}-a b-a b+b^{2}\right)(a-b)} \\ \\ {=\left(a^{2}-2 a b+b^{2}\right)(a-b)}\end{array}$

Multiplying two terms, we get

$=\left[a^{2}(a-b)-2 a b(a-b)+b^{2}(a-b)\right]$

$=\left(a^{3}-a^{2} b-2 a^{2} b+2 a b^{2}+a b^{2}-b^{3}\right)$

$=\left(a^{3}-3 a^{2} b+3 a b^{2}-b^{3}\right)$