Question

39. Find the value of
(i) x3 + y3 -- 12xy + 64, when x + y = -4
(ii) x3 -- 8y3 – 36xy – 216, when x = 2y + 6​

Answer 1

1. x³+y³-12xy+64

x=-4-y

(-4-y)³ + y³-12(-4-y)+64

-(64+y³+48y+12y²)+y³+48+12y+64

-64-y³-48y-12y²+y³+48+12y+64

12y²+36y-48=0

12(y²+3y-4)

12(y²+4y-y-4)

12(y(y+4)-1(y+4))

12(y-1)(y+4)

y=1 y= -4

put the values of y in 1st equation nd find the answer

Answer 2

Step-by-step explanation:

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