The standard form of $y = (2x - 1)^3 - (3x-3)^2$ ?

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Answer 1 · 2018-01-26T10:20:28

$8x^3-21x^2+24x-10$

First expand $(2x-1)^3$ :

$(2x)^3+3(2x)^2(-1)+3(2x)(-1)^2+(-1)^3$

$=8x^3-3(4x^2)+6x-1$

$=8x^3-12x^2+6x-1$

expanding $(3x-3)^2$ :

$9x^2-18x+9$

Substituting in the original:

$(2x-1)^3-(3x-3)^2=$

$=(8x^3-12x^2+6x-1)-(9x^2-18x+9)$

$=8x^3-12x^2+6x-1 - 9x^2+18x-9$

$=8x^3-21x^2+24x-10$

The standard form of y = (2x - 1)^3 ­- (3x-3)^2y = (2x - 1)^3 ­- (3x-3)^2?