Question

Prove that (a^3)(b^3)=(a^3)+(b^3)+3ab(a+b)

Answer 1

Step-by-step explanation:

I think it isn't a³b³. It will (a+b)³=a³+b³+3ab(a+b).

For this, take the L.H.S.

now,L.H.S.=(a+b)³

=(a+b)²(a+b)

=(a²+b²+2ab)(a+b)

=a(a²+b²+2ab)+b(a²+b²+2ab)

=a³+ab²+2a²b+a²b+b³+2ab²

=a³+b³+3a²b+3ab²

=a³+b³+3ab(a+b)

...... proved

Answer 2

$\Large{\underline{\underline{\mathfrak{To \: Prove:-}}}}$

$\rm{(a+b)^3 = a^3 + b^3 + 3ab(a+b) }$

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$\Large{\underline{\underline{\mathfrak{Solution:-}}}}$

$\rm{we \: have,}$

$\rm{(a+b)^3=(a+b)(a+b)(a+b)}$

${\bold{\underline{\underline{[By\: the \: law \: of \: exponents]}}}}$

$\rm{(a+b)^3=(a+b)(a+b)^2}$

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

$\rm{(a+b)^3=a×(a^2+2ab+b^2)+b×(a^2+2ab+b^2}$

${\bold{\underline{\underline{[By\: distributive\: law]}}}}$

$\rm{(a+b)^3=a×a^2+a×2ab +a×b^2+b×a^2+b×2ab+b×b^2}$

${\bold{\underline{\underline{[By\: distributive\: law]}}}}$

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

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

${\bold{\underline{\underline{[Rearranging\: the \:terms]}}}}$

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

$\rm{(a+b)^3=a^3+b^3+3ab(a+b)}$

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$\Large{\underline{\underline{\mathfrak{Hence \: Proved!}}}}$