Question

plz do this fast
for 9th class students

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Prove that

$\rm \: \dfrac{0.87 \times 0.87 \times 0.7 + 0.13 \times 0.13 \times 0.13}{0.87 \times 0.87 - 0.87 \times 0.13 + 0.13 \times 0.13} = 1$

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

Consider, LHS

$\rm :\longmapsto\: \: \dfrac{0.87 \times 0.87 \times 0.7 + 0.13 \times 0.13 \times 0.13}{0.87 \times 0.87 - 0.87 \times 0.13 + 0.13 \times 0.13}$

We know,

${ \bf{ \underbrace{ \sf \: x \times x \times x \times - - - \times x} \: = {x}^{n} }} \\ \rm \: n \: times$

So, using this identity, we get

$\rm \:  =  \:  \: \dfrac{ {(0.87)}^{3} + {(0.13)}^{3} }{ {(0.87)}^{2} - 0.87 \times 0.13 + {(0.13)}^{2} }$

We know, that,

$\boxed{ \bf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})}}$

So, using this identity in numerator, we get

$\rm \:  =  \:  \: \dfrac{ {(0.87}^{} + {0.13)}^{} ({(0.87)}^{2} - 0.87 \times 0.13 + {(0.13)}^{2})}{ {(0.87)}^{2} - 0.87 \times 0.13 + {(0.13)}^{2} }$

$\rm \:  =  \:  \: 0.87 + 0.13$

$\rm \:  =  \:  \: 1$

Hence,

$\boxed{ \bf{ \: \rm \: \dfrac{0.87 \times 0.87 \times 0.7 + 0.13 \times 0.13 \times 0.13}{0.87 \times 0.87 - 0.87 \times 0.13 + 0.13 \times 0.13} = 1}}$

Short Cut trick :-

To solve such questions, use these Identities directly.

1.

$\boxed{ \bf{ \: \dfrac{ {x}^{3} + {y}^{3} }{ {x}^{2} - xy + {y}^{2}} = x + y}}$

So, using this identity,

$\rm :\longmapsto\: \: \dfrac{0.87 \times 0.87 \times 0.7 + 0.13 \times 0.13 \times 0.13}{0.87 \times 0.87 - 0.87 \times 0.13 + 0.13 \times 0.13}$

$\rm \:  =  \:  \: 0.87 + 0.13$

$\rm \:  =  \:  \: 1$

Additional Information :-

$\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}$

$\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$\boxed{ \bf{ \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2})}}$