Question

...............................HELP ME​

Answer 1

Answer -

• refer the attachment

Hope it will help you

Answer 2

$\tt{i)\ x^3+3x^2+3x+28=x^3+3x^2+3x+1+27}\\ \\ \implies \tt{(x+1)^3+(3)^3}\\ \\ \tt{We\ know\ that\ :} \ a^3 + b^3=(a+b)(a^+b^2-ab) \\ \\ \tt{Using\ this\ identity\ we\ simplify\ this\ further\ :}\\ \\ \implies \tt{(x+1+3)\Big((x+1)^2+3^2-3(x+1)\Big)} \\ \\ \implies \tt{(x+4)(x^2-x+7)}$

$\tt{ii)\ x^3 - 3x-1+\dfrac{3}{x}-\dfrac{1}{x^3}} \\ \\ \tt{Rearranging\ the\ terms:}\\ \\ \implies\Big(x^3-\dfrac{1}{x^3}\Big)-3\Big(x-\dfrac{1}{x}\Big)-1\\ \\ \implies \Big(x-\dfrac{1}{x}\Big)\Big(x^2+\dfrac{1}{x^2}-1\Big)-3\Big(x-\dfrac{1}{x}\Big)-1\\ \\ Taking\ \Big(x-\dfrac{1}{x}\Big) \ common\ in\ first \ two\ terms: \\ \\ \implies \Big(x-\dfrac{1}{x}\Big)\Big(x^2+\dfrac{1}{x^2}+1-3\Big)-1\\ \\ \implies \Big(x-\dfrac{1}{x}\Big)\Big(x^2+\dfrac{1}{x^2}-2\Big)-1$

$\tt \Big(x-\dfrac{1}{x}\Big)\Big(x^2+\dfrac{1}{x^2}-2(x^2)\Big(\dfrac{1}{x^2}\Big)\Big)-1 \\ \\ \implies \Big(x-\dfrac{1}{x}\Big)\Big(x-\dfrac{1}{x}\Big)^2-1 \\ \\ \implies \Big(x-\dfrac{1}{x}\Big)^3-1^3\\ \\ \implies \Big(x-\dfrac{1}{x}-1\Big)\Big(x^2+\dfrac{1}{x^2}+x-\dfrac{1}{x}-1\Big)$

Hope this helps.