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Question
If a, b, c and d are in proportion, prove that: `(a^2 + ab + b^2)/(a^2 - ab + b^2) = (c^2 + cd + d^2)/(c^2 - cd + d^2)`
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Solution
∵ a, b, c, d are in proportion
`a/b = c/d` = k(say)
a = bk, c = dk.
L.H.S. = `(a^2 + ab + b^2)/(a^2 - ab + b^2)`
= `(b^2k^2 + bk.b + b^2)/(b^2k^2 - bk.b + b^2)`
= `(b^2(k^2 + k + 1))/(b^2(k^2 - k + 1)`
= `(k^2 + k + 1)/(k^2 - k + 1)`
R.H.S. = `(c^2 + cd + d^2)/(c^2 - cd + d^2)`
= `(d^2k^2 + dkd + d^2)/(d^2k^2 - dk.d + d^2)`
= `(d^2(k^2 + k + 1))/(d^2(k^2 - k + 1)`
= `(k^2 + k + 1)/(k^2 - k + 1)`
∴ L.H.S. = R.H.S.
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