Question

evaluate a cube - b cube when a-b =2 and a square + b square = 4​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: {a}^{2} + {b}^{2} = 4 - - - (1)$

and

$\rm \: a - b = 2 - - - (2)$

On squaring both sides, we get

$\rm \: {(a - b)}^{2} = {2}^{2}$

$\rm \: {a}^{2} + {b}^{2} - 2ab = 4$

$\rm \: 4 - 2ab = 4 \: \: \: \{using \: (1) \}$

$\rm \: - 2ab = 0$

$\rm\implies \:ab = 0$

Now, Consider

$\rm \: {a}^{3} - {b}^{3}$

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

On substituting the values, we get

$\rm \: =  \:(2)(4 + 0)$

$\rm \: =  \:2 \times 4$

$\rm \: =  \:8$

Hence,

$\rm\implies \:\boxed{\rm{  \: {a}^{3} - {b}^{3} = 8 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$