Question 1 Report
If ac a c = cd c d = k, find the value of 3a2?ac+c23b2?bd+d2
Answer Details
ac a c = cd c d = k ∴ ab a b = bk cd c d = k ∴ c = dk = 3a2−ac+c23b2−bd+d2 3 a 2 − a c + c 2 3 b 2 − b d + d 2 = 3(bk)2−(bk)(dk)+dk23b2−bd+a2 3 ( b k ) 2 − ( b k ) ( d k ) + d k 2 3 b 2 − b d + a 2 = 3b2k2−bk2d+dk23b2−bd+d2 3 b 2 k 2 − b k 2 d + d k 2 3 b 2 − b d + d 2 k = 3b2k2−bk2d+dk23b2−bd+d2
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