Asked by Meera
If 3x+2y=12 and xy=6,find 27x^3+8y^3
Answers
Answered by
Reiny
27x^3+8y^3 --- the difference of cubes
= (3x + 2y)(9x^2 - 6xy + 4y^2)
= 12(9x^2 - 6xy + 4y^2)
but 9x^2 - 6xy + 4y^2
= (9x^2 + 12xy + 4y^2) - 18xy
= (3x+2y)^2 - 18xy
= 144 - 108
= 36
so 27x^3 + 8y^3 = 12(36) = 4320
or:
y = 6/x
back in the first:
3x + 2(6/x) = 12
times x
3x^2 + 12 = 12x
3x^2 - 12x + 12 = 0
x^2 - 4x + 4 = 0
(x-2)^2 = 0
x-2 = 0
x = 2
y = 6/2 = 3
then 27x^3+8y^3
= 27(8) + 8(27) = 432
= (3x + 2y)(9x^2 - 6xy + 4y^2)
= 12(9x^2 - 6xy + 4y^2)
but 9x^2 - 6xy + 4y^2
= (9x^2 + 12xy + 4y^2) - 18xy
= (3x+2y)^2 - 18xy
= 144 - 108
= 36
so 27x^3 + 8y^3 = 12(36) = 4320
or:
y = 6/x
back in the first:
3x + 2(6/x) = 12
times x
3x^2 + 12 = 12x
3x^2 - 12x + 12 = 0
x^2 - 4x + 4 = 0
(x-2)^2 = 0
x-2 = 0
x = 2
y = 6/2 = 3
then 27x^3+8y^3
= 27(8) + 8(27) = 432
Answered by
Anonymous
Simple one
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