Question

verify that a cube plus b cube equal to a + b whole into a square minus a b + b square

Answer 1

$<b>$Answer :

$to \: prove = > {a}^{3} + {b}^{3} = (a + b)( {a}^{2} + {b}^{2} - ab) \\ \\ = > {a}^{3} + {b}^{ 3} = a( {a}^{2} + {b}^{2} - ab) + b( {a}^{2} + {b}^{2} - ab)\\ \\ = > {a}^{ 3} + {b}^{3} = a \times {a}^{2} + a \times {b}^{2} + a \times - ab + b \times {a}^{2} + b \times {b}^{2} + b \times - ab \\ \\ = > {a}^{3} + {b }^{ 3} = {a}^{3} + a {b}^{2} - {a}^{2} b + {a}^{2} b + {b}^{3} - a {b}^{2} \\ \\ cancelling \: the \: similar\: terms \: with \: opposite \: signs \\ \\ = > {a}^{3} + {b}^{3} = {a}^{3} + {b}^{3}$



$\color{brown}\underline\textbf{Hence, proved. }$

Tysm for the question!

Answer 2

We know that,

(a+b)³ = (a+b)²(a+b)

(a+b)³ = (a² + b² + 2ab)(a+b)

(a+b)³ = a(a² + b² + 2ab) +b(a² + b² + 2ab)

(a+b)³ = a³ + ab² + 2a²b + a²b + b³ +2ab²

(a+b)³ = a³ + b³ + 3ab² + 3a²b

We have find a³ + b³. therefore, transfer other terms to one side:

(a+b)³ - 3a²b - 3a²b = a³ + b³

a³ + b³ = (a+b)³ - 3a²b - 3a²b

a³ + b³ = (a+b)³ -3ab(a+b)

Taking (a+b) common,

a³ + b³ = (a+b){(a+b)² -3ab}

a³ + b³ = (a+b){a² + b² + 2ab -3ab}

a³ + b³ = (a+b)(a² + b² - ab)

Hence proved;

Note:-
(a+b)² = (a+b)(a+b)
= a(a+b) +b(a+b)
= a² + ab + ba + b²
= a² + b² + 2ab