Question

factorize a^3+b^3+3ab-1

Answer 1

Heya!!

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= [a^3 + b^3 - 1^3 - 3(a)(b)(-1)]

By identity of x^3 + y^3 +z^3 - 3xyz;

= (a+b-1) (a^2 + b^2 +1 - ab+a+b)

Hope it helps uh :)

Answer 2

$\textbf{\underline{\underline{Answer :}}}$

We know that,

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Now, a³ + b³ + 3ab - 1

= a³ + b³ - 1 + 3ab

= a³ + b³ + (- 1)³ - 3ab(- 1)

= (a + b - 1) {a² + b² + (- 1)² - ab - b(- 1) - (- 1)a}

= (a + b - 1) (a² + b² + 1 - ab + b + a),

which is the required factorization.

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