Question

if 5a+3b =7, a+1/b =0 then ( 125a³+27b³ ) =?​

Answer 1

Answer:

given that

a+1/b=0 then a.b= –1

5a+3b=7 (cube on both side)

(5a+3b)³=7³

125a³+27b³+3×5a×3b(5a+3b)=343 :: { (a+b)³= a³+b³+3ab(a+b) }

125a³+27b³–45×7=343. {::a×b= –1, 5a+3b=7}

125a³+27b³=343+315

125a³+27b³= 658 solved

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