If \(\frac{a}{c}\) = \(\frac{c}{d}\) = k, find the value of \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\)

3k²

2k - k²

\(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)

k²

Correct answer is C

\(\frac{a}{c}\) = \(\frac{c}{d}\) = k

∴ \(\frac{a}{b}\) = bk

\(\frac{c}{d}\) = k

∴ c = dk

= \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\)

= \(\frac{3(bk)^2 - (bk)(dk) + dk^2}{3b^2 - bd + a^2}\)

= \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)

k = \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)

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If \(\frac{a}{c}\) = \(\frac{c}{d}\) = k, find the value of \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\)

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