Math, asked by delhimanojshukla, 1 year ago

Find the value of x³-8y³-36xy-216when x=2y+6

Answers

Answered by AvikaJaiswal
0

Answer:

x³-8y³-36xy-216 when x=2y +6 then we take the place of x is x value (2y+6)³-8y³-36xy(2u+6)y-216

8y+18-8y³-72xy²+216-216=-8y³+8y+18-72xy²=-8y³-72xy²38y+18

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