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In case you meant 64x^3 - 27y^3, that factors to (4x - 3y)(16x^2 + 12xy + 9y^2)

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There is no factoring required - the answer is 37x3

Q: Factor 64x to the third power minus 27x to the third power?

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(5x - 2)(x - 5)

9x2 + 27x - 36 = 9(x2 + 3x - 4) = 9(x + 4)(x - 1)

You try to find two factors of -58 whose sum is -27 .You find that -29 and +2 will do it, so . . .x2 - 27x - 58 = (x - 29) (x + 2)

3x2 + 27x +60

9xy(3xy - 4)

Related questions

-x^6(64x^3 - 27)

(2 - 3x)(9x^2 + 6x + 4)

(5x - 2)(x - 5)

9x2 + 27x - 36 = 9(x2 + 3x - 4) = 9(x + 4)(x - 1)

You try to find two factors of -58 whose sum is -27 .You find that -29 and +2 will do it, so . . .x2 - 27x - 58 = (x - 29) (x + 2)

20x2 - 27x -8 20 * -8 is -160, and -32 and 5 add to -27 20x2 - 32x + 5x - 8 4x(5x-8) + 1(5x-8) (4x + 1)(5x-8)

3x2 + 27x +60

27X^3-8 = 0 27X^3 = 8 X^3 = 8/27 X = cubic root of 8/cubic root of 27 = 2/3

27x + 3 use distibutive law to factor

0

(2x - 3)(x - 12)

9xy(3xy - 4)