Question

prove that (a-b)³=a³-b³-3a²b+3a²b​

Answer 1

Correct Question :–

→ Prove that (a-b)³=a³-b³-3a²b+3ab².

ANSWER :–

TO PROVE :–

$\\ \bf \longrightarrow {(a - b)}^{3} = {a}^{3} - {b}^{3} - 3 {a}^{2} b + 3a {b}^{2}$

SOLUTION :–

• Let's take L.H.S. –

$\\ \bf \: \:= \: \: {(a - b)}^{3}$

• We should write this as –

$\\ \bf \: \:= \: \:(a - b)(a - b)(a - b)$

$\\ \bf \: \:= \: \:\underbrace{[(a - b)(a - b)]}(a - b)$

$\\ \bf \: \:= \: \:( {a}^{2} - ab - ab + {b}^{2})(a - b)$

$\\ \bf \: \:= \: \:( {a}^{2} - 2ab +{b}^{2})(a - b)$

$\\ \bf \: \:= \: \:(a - b)( {a}^{2} - 2ab +{b}^{2})$

$\\ \bf \: \:= \: \:a( {a}^{2} - 2ab +{b}^{2})- b({a}^{2} - 2ab +{b}^{2})$

$\\ \bf \: \:= \: \: {a}^{3} - 2 {a}^{2}b + a {b}^{2} - {a}^{2}b + 2a {b}^{2} - {b}^{3}$

$\\ \bf \: \:= \: \: {a}^{3} - 3 {a}^{2}b + 3a {b}^{2}- {b}^{3}$

$\\ \bf \: \:= \: \: {a}^{3}- {b}^{3} - 3 {a}^{2}b + 3a {b}^{2}$

$\\ \bf \: \:= \: \: R.H.S.$

$\\ \large \longrightarrow { \boxed{\bf Hence \: \: Proved}}$

Answer 2

Correct question

★ Prove that (a - b)³ = a³ - b³ - 3a²b + 3ab²

Solution

Taking L.H.S

→ (a - b)³

→ (a - b)(a - b)(a - b)

→ [(a - b)(a - b)](a - b)

→ [a² - ab - ab + b²](a - b)

→ (a - b)(a² -2ab + b²)

→ a × (a² -2ab + b²) - b × (a² -2ab + b²)

→ a³ - 2a²b + ab² - ba² + 2ab² - b³

→ a³ - b³ - (2a²b + a²b) + (2ab² + ab²)

→ a³ - b³ - 3a²b + 3ab²

⟹ R.H.S

⟹ L.H.S = R.H.S

Hence proved