Question

EX Pand:(x+2)(x-1)(x-3)​

Answer 1

$\huge\color{red}{\bf{Answer:}}$

$\bf (x + 2)(x - 1)(x - 3) \\ \bf = [ \: (x \: \times \: x ) + (x \: \times - 1) + (2 \times \: x) + (2 \times - 1) \: ](x - 3) \\ \bf = ( {x}^{2} - x + 2x - 2)(x - 3) \\ \bf = (x - 3) ({x}^{2} + x - 2) \\ \bf = (x \: \times \: {x}^{2} ) + (x \: \times \: x) + (x \: \times - 2) + ( - 3 \times {x}^{2} ) + ( - 3 \times \: x) + ( - 3 \times - 2) \\ \bf = {x}^{3} + {x}^{2} - 2x - {3x}^{2} - 3x + 6 \\ \bf = {x}^{3} + ( {x}^{2} - {3x}^{2} ) + ( - 2x - 3x) + 6 \\ \bf = \underline{ \underline{{x}^{3} - {2x}^{2} - 5x + 6}}$

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Answer 2

$\sf \: We \: know \: that (x+y)3=x3+y3+3xy(x+y)$

$\sf \: Here \: x=x \: and y=1 we \: get$

$\sf(x+1)3=x3+(1)3+3(x)(1)(x+1)$

$\sf=x3+1+3x(x+1)$

$\sf=x3+1+3x2+3x$