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What is the standard form of $y = {(- x + 1)}^{3} - {(- 3 x + 1)}^{2}$ $y= (-x+1)^3-(-3x+1)^2$ ?

1 Answer

Shiva Prakash M V

Mar 29, 2018

$y = - 28 x^{3} + 30 x^{2} - 12 x + 2$ $y=-28x^3+30x^2-12x+2$

Explanation:

$y = {(- x + 1)}^{3} - {(- 3 x + 1)}^{2}$ $y=(-x+1)^3-(-3x+1)^2$

$y = {(1 - x)}^{3} + {(1 - 3 x)}^{2}$ $y=(1-x)^3+(1-3x)^2$

$y = 1^{3} - 3 \times 1^{2} \times x + 3 \times 1 \times x^{2} - x^{3} + 1^{3} - 3 \times 1^{2} \times 3 x + 3 \times 1 \times {(3 x)}^{2} - {(3 x)}^{3}$ $y=1^3-3xx1^2 xxx+3xx1xxx^2-x^3+1^3-3xx1^2xx3x+3xx1xx(3x)^2-(3x)^3$

$y = 1 - 3 x + 3 x^{2} - x^{3} + 1 - 9 x + 27 x^{2} - 27 x^{3}$ $y=1-3x+3x^2-x^3+1-9x+27x^2-27x^3$

$y = - x^{3} - 27 x^{3} + 3 x^{2} + 27 x^{2} - 9 x - 3 x + 1 + 1$ $y=-x^3-27x^3+3x^2+27x^2-9x-3x+1+1$

$y = - 28 x^{3} + 30 x^{2} - 12 x + 2$ $y=-28x^3+30x^2-12x+2$

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