Question

pls solve it fast it is of math algnra

Answer 1

Given :-

$\sf{ \implies \: x \: = \sqrt[3]{7 + 5 \sqrt{2} } - \frac{1}{ \sqrt[3]{7 + 5 \sqrt{2} } }}$

To find :-

$\sf{Value\: of \: {x}^{3} + 3x 11}$

Solution :-

$\sf{ \implies \: x \: = \sqrt[3]{7 + 5 \sqrt{2} } - \frac{1}{ \sqrt[3]{7 + 5 \sqrt{2} } }}$

Cubing both sides

$\sf{ \implies \: {x}^{3} \: = {\sqrt[3]{7 + 5 \sqrt{2} } - \frac{1}{ \sqrt[3]{7 + 5 \sqrt{2} } }}^{3}}$

It resembles with the identity of (a-b)³

$\sf{\implies {(a - b)}^{3} = {a}^{3} - {b}^{3} - 3ab( a - b) }$

Applying this identity over here .

$\sf{\implies \: ( { \sqrt[3]{7 + 5 \sqrt{2} } })^{3} - ( { \frac{1}{ \sqrt[3]{7 + 5 \sqrt{2} } } )}^{3} - 3( \sqrt[3]{7 + 5 \sqrt{2} } \times \frac{1}{ \sqrt[3]{7 + 5 \sqrt{2} } } )( \sqrt[3]{7 + 5 \sqrt{2} - \frac{1}{ \sqrt[3]{7 + 5 \sqrt{2} } } } }$

Now we know that $\sf{ \implies \: x \: = \sqrt[3]{7 + 5 \sqrt{2} } - \frac{1}{ \sqrt[3]{7 + 5 \sqrt{2} } }}$

Replacing x for its value.

$\sf{\implies 7 + 5 \sqrt{2} - \frac{1}{7 + 5 \sqrt{2} } - 3 (x) }$

Now rationalising $\sf{ \frac{1}{7 + 5 \sqrt{2}}}$

$\sf {\implies \: \frac{1}{7 + 5 \sqrt{2} } \times \frac{7 - 5 \sqrt{2} }{7 - 5 \sqrt{2} } }$

$\sf{ \implies \frac{ 7 - 5 \sqrt{2}}{ {7}^{2} - {5 \sqrt{2}}^{2}}}$

$\sf{\implies \frac{ 7 - 5 \sqrt{2} }{ 49 - 50} }$

$\sf{\implies \frac{ 7 - 5 \sqrt{2} }{ -1 } }$

$\sf{\implies 5 \sqrt{2} - 7}$

Now putting this value in simplification of identity .

$\sf{\implies 7 + 5 \sqrt{2} - ( 5 \sqrt{2} - 7 ) - 3x }$

$\sf{\implies 7 + 5 \sqrt{2} - 5 \sqrt{2} + 7 - 3x }$

${\underline{\underline{\sf{\implies {x}^{3} = 14 - 3x }}}}$

Putting the value of x³ in x³ + 3x - 11 .

$\sf{\implies 14 - 3x + 3x - 11 }$

$\sf{\implies 14 - 11 = 3 }$

${\underline{\underline{\sf{\implies value \: of \: expression \: is \: 3 }}}}$