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Question
If x, y, z are in continued proportion, prove that: xyz(x + y + z)3 = (xy + yz + zx)3.
Theorem
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Solution
x, y, z are in continued proportion,
`x/y = y/z` = k
y = zk
x = yk = zk2
L.H.S.
= xyz(x + y + z)3
= (zk2)(zk)(z)(zk2 + zk + z)3
= (z3k3)[z(k2 + k + 1)]3
= z3k3 . z3(k2 + k + 1)3
= z6k3(k2 + k + 1)3
R.H.S.
= (xy + yz + zx)3
= (zk2 ⋅ zk + zk ⋅ z + z ⋅ zk2)3
= (z2k3 + z2k + z2k2)3
= [z2(k3 + k2 + k)]3
= z6(k3 + k2 + k)3
= z6[k(k2 + k + 1)]3
= z6k3(k2 + k + 1)3
L.H.S. = R.H.S.
Hence proved.
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Chapter 7: Ratio and proportion - Exercise 7B [Page 126]
