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Question

If 9xcosθ+5ysinθ=56 and 9xsinθcos2θ5ycosθsin2θ=0, then the value of [(9x)2/3+(5y)2/3]3 is

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Solution

Given, 9xcosθ+5ysinθ=56 .....(1)

Also given, 9xsinθcos2θ5ycosθsin2θ=0

9xsin3θ=5ycos3θ.

cos3θ9x=sin3θ5y=k3 (say)

cosθ=k(9x)1/3 and sinθ=k(5y)1/3 ....(2)

Squaring and adding, we get
1=cos2θ+sin2θ
k2[(9x)2/3+(5y)2/3]=1
k2=1[(9x)2/3+(5y)2/3]
Substitute the value from (2) in (1), we get
9xk(9x)1/3+5yk(5y)1/3=56 (by (1))

(9x)2/3+(5y)2/3=56k

[(9x)2/3+(5y)2/3]2=(56)2k2

[(9x)2/3+(5y)2/3]2=(56)2(9x)2/3+(5y)2/3

[(9x)2/3+(5y)2/3]3=(56)2=3136.

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