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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:-
[tex]a) \frac{p}{ {q}^{3} } \times ( {p}^{2} - 3q)[/tex]
Given,
x + y = p
and xy = q .
To find:-
[tex] \frac{1}{ {x}^{3} } + \frac{1}{ {y}^{3} } [/tex]
Solution:-
[tex] \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} } [/tex]
[tex] \frac{p}{q} \times ( {p}^{2} - 3q)[/tex]
Hope it's helpful to you.
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