tricky approach (Question: 259989.0 )

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prakash deep

· started a discussion

· 1 Months ago

tricky approach (Question: 259989.0 )

let x/y = a then y/x = 1/a
now a-1/a= -3
a+1/a= underroot minus 3 square plus 4=root 13 = A (let)
now; a cube plus 1/a cube= A CUBE minus 3A(FORMULA) =10root 13

Question:

If $\frac{y}{x} - \frac{x}{y} = 3,$ , then find the value of $\frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}}$ is-

Options:

10 $\sqrt{3}$

10 $\sqrt{13}$

13 $\sqrt{3}$

13 $\sqrt{10}$

Solution:

Ans: (b)

priyanka

· commented

· 1 Months ago

nice approach

Savita

· commented

· 1 Months ago

Thnx brother, it really helped......

BRIJESH PRAJAPATI

· commented

· 1 Months ago

Thank you brother

manjeet kumar

· commented

· 1 Months ago

thanks bhai

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tricky approach (Question: 259989.0 )

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