Question

Please solve this step-by-step.

Answer 1

Answer:

x^2 + 1/x^2 - 2/x - 2x + 3

Step-by-step explanation:

Note :

( a+b+c ) ^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

given question is in the form of ( a+b+c )^2

where a = x

b = 1/x

c = - 1

[ x+1/x+(-1) ]^2 = x^2 + (1/x)^2 + (-1)^2 + 2 (x)(1/x) +2 (1/x)(-1) + 2 (-1)(x)

=> [ x+1/x+(-1) ]^2 = x^2 + 1/x^2 + 1 + 2 - 2/x - 2x

=> ( x + 1/x - 1 ) ^2 = x^2 + 1/x^2 - 2/x - 2x + 3

Answer 2

$\sf{(x\:+\dfrac{1}{x}\:-1)^2}$

This question is in the form – (a + b + c)²

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

In the same way, expanding this equation:

$\sf{(x+\dfrac{1}{x}-1)^2}$

$\sf{=x^2+\dfrac{1}{x}^2+(- 1)^2+(2\:\times\:x\:\times\dfrac{1}{x})+(2\:\times\dfrac{1}{x}\times-1)+2\times-1\times\:x}$

$\sf{=x^2+\dfrac{1}{x^2}+1+\dfrac{2}{x}-\dfrac{2}{x}-2x}$

$\sf{=-2x+x^2+\dfrac{1}{x^2}+1}$

Therefore, we get $\sf{-2x+x^2+\dfrac{1}{x^2}+1}$ after simplifying $\sf{(x\:+\dfrac{1}{x}\:-1)^2}$.

More Algebraic Identities :

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$