Question

If x is real and x^3+1/x^3=0 then find the value of (x+1/x)^2​

Answer 1

Answer:

what is the

Step-by-step explanation:

chapter name?

Answer 2

$\bf \LARGE\color{pink}Hello!$

${ \huge{ \star}} \: \: \: \: { \underline{\bf \LARGE Given :}}$

$\rightsquigarrow \sf \: x \in \: R$

$\rightsquigarrow \sf \: {x}^{3} + \frac{1}{ {x}^{3} } = 0$

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${ \huge{ \star}} \: \: \: \: { \underline{\bf \LARGE To \: \: Find :}}$

$\sf \rightsquigarrow \: {(x + \frac{1}{x} )}^{2} = {?}$

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${ \huge{ \star}} \: \: \: \: { \underline{\bf \LARGE Solution :}}$

$\tt \fcolorbox{skyblue}{skyblue}{Process - 1}$

$\: \: \circ \: \: \: \: \: \: \: \underline{\sf \: {x}^{3} + \frac{1}{ {x}^{3} } = 0 }$

$\implies \sf (x + \frac{1}{x} )( {x}^{2} - x \times \frac{1}{x} + \frac{1}{ {x}^{2} } ) = 0\: \: \: \: \bf[Formula]$

$\implies \sf \: (x + \frac{1}{x} ) = 0{ \underline{ \: \: \: \: \: \: \: \: \: \: }}(1)$

$\: \: \therefore \: \: \: \: \: \: { \underline{ \boxed{ \sf \: (x + \frac{1}{x} ) ^{2} = 0}}}$

$\tt \fcolorbox{skyblue}{skyblue}{Process - 2}$

$\: \: \circ \: \: \: \: \: \: \: \underline{\sf \: {x}^{3} + \frac{1}{ {x}^{3} } = 0 }$

$\implies \sf \: {(x + \frac{1}{x}) }^{3} - 3 \: x \frac{1}{x} (x + \frac{1}{x} )\: \: \: \: \bf[Formula]$

From eqn (1)

$\implies \sf(x + \frac{1}{x} ) {}^{3} = 0 \: \: \: \: \: \bf[\because (x + \frac{1}{x} ) = 0]$

$\: \: \therefore \: \: \: \: \: \: { \underline{ \boxed{ \sf \: (x + \frac{1}{x} ) ^{2} = 0}}}$

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${ \huge{ \star}} \: \: \: \: { \underline{\bf \LARGE Formula \: \: Required:}}$

$\sf\rightsquigarrow \: {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )$

$\sf\rightsquigarrow \: {a}^{3} + {b}^{3} = (a + b) {}^{3} - 3ab(a + b)$

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HOPE THIS IS HELPFUL...

$\tt \fcolorbox{skyblue}{skyblue}{@StayHigh}$