Question

write the formula a³- b³ pls answer​

Answer 1

Step-by-step explanation:

a³-b³= (a-b)(a²+ab+ b²)

Answer 2

$\begin{gathered}{\underline{\underbrace{\Huge{\bigstar{\textsf{\textbf{\blue{Solution :}}}}}}}}\end{gathered}$

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$\red{\large \qquad \boxed{\boxed{\begin{array}{cc} \tt a³- b³ \\ \\ \tt a³-b³= (a-b)(a²+ab+ b²)\end{array}}}}$

More Explanation:-

$\footnotesize{ \tt (i) a³+b³ = (a+b)(a²-ab+b²)}$

$\tt Or$

$\footnotesize{ \tt {= (a+b)³-3ab(a+b)}}$

$\footnotesize \tt{ii) a³-b³ = (a-b)(a²+ab+b²)}$

$\tt Or$

$\footnotesize{ \tt{= (a-b)³+3ab(a-b)}}$

$\boxed{\tt{i) We \: know \: the \: algebraic \: identity:}}$

$\footnotesize \tt{a³+3a²b+3ab²+b³ = (a+b)³}$

$\footnotesize \tt{ \implies{ a³+b³+3ab(a+b)=(a+b)³}}$

$\footnotesize{ \tt \implies{a³+b³ = (a+b)³-3ab(a+b) \longrightarrow (1)}}$

$\footnotesize \tt{(a+b)[(a+b)²-3ab]}$

$\footnotesize \tt{ (a+b)(a²+2ab+b²-3ab)}$

$\footnotesize{ \tt{ \implies(a+b)(a²-ab+b²) \longrightarrow(2)}}$

$\boxed {\tt ii) By \: algebraic \: identity:,}$

$\footnotesize \tt{a³-3a²b+3ab²-b³ = (a-b)³}$

$\footnotesize \tt{\implies a³-b³-3ab(a-b)=(a-b)³}$

$\footnotesize \tt{\implies a³-b³ = (a-b)³+3ab(a-b) \longrightarrow(3)}$

$\footnotesize \tt{ \implies (a-b)[(a-b)²+3ab]}$

$\footnotesize \tt{ \implies(a-b)(a²-2ab+b²+3ab)}$

$\footnotesize \tt{ \implies (a-b)(a²+ab+b²) \longrightarrow(4)}$

$\footnotesize \tt{i)a³+b³ = (a+b)(a²-ab+b²)}$

$\tt Or$

$\footnotesize \tt{ \implies (a+b)³-3ab(a+b)}$

$\footnotesize \tt{ii) a³-b³ =(a-b)(a²+ab+b²)}$

$\tt Or$

$\footnotesize \tt{ \implies (a-b)³+3ab(a-b)}$

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