Question

Solve this question No spamming​

Answer 1

Answer:

(a)

Step-by-step explanation:

 ⇒ x + y = p

⇒ ( x + y )^3 = p^3

      ( x + y )^3 = x^3 + y^3 + 3xy( x + y )

⇒ x^3 + y^3 + 3xy( x + y ) = p^3

        xy = p  ;  x + y = p

⇒ x^3 + y^3 + 3q(p) = p^3

⇒ x^3 + y^3 = p^3 - 3pq

        In question :

⇒ 1/x^3 + 1/y^3

⇒ ( y^3 + x^3 ) / ( xy )^3

       x^3 + y^3 = p^3 - 3pq

            xy = q

⇒ ( p^3 - 3pq ) / q^3

⇒ p( p^2 - 3q ) / q^3

    Option (a)

Answer 2

Answer:-

$a) \frac{p}{ {q}^{3} } \times ( {p}^{2} - 3q)$

Given,

x + y = p

and xy = q .

To find:-

$\frac{1}{ {x}^{3} } + \frac{1}{ {y}^{3} }$

Solution:-

$\frac{1}{ {x}^{3} } + \frac{1}{ {y}^{3} } \\ = \frac{ {y}^{3} + {x}^{3} }{ {x}^{3} \times {y}^{3} } \\ = \frac{ {(x + y)}^{3} - 3xy \times (x + y)}{ {(xy)}^{3} } \\ = \frac{ {p}^{3} - 3pq }{ {q}^{3} }$

$\frac{p}{q} \times ( {p}^{2} - 3q)$

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