Question

If x-1/x =4 , find the value of x^3-1/x^3​

Answer 1

EXPLANATION.

⇒ (x - 1/x) = 4.

As we know that,

Cubing on both sides of the equation, we get.

⇒ (x - 1/x)³ = (4)³.

As we know that,

Formula of :

⇒ (a - b)³ = a³ - 3a²b + 3ab² - b³.

Using this formula in the equation, we get.

⇒ (x)³ - 3(x)²(1/x) + 3(x)(1/x)² - (1/x)³ = (4)³.

⇒ x³ - 3x + 3/x - 1/x³ = 64.

⇒ x³ - 3(x - 1/x) - 1/x³ = 64.

Put the value of (x - 1/x) = 4 in the equation, we get.

⇒ x³ - 1/x³ - 3(x - 1/x) = 64.

⇒ x³ - 1/x³ - 3(4) = 64.

⇒ x³ - 1/x³ - 12 = 64.

⇒ x³ - 1/x³ = 64 + 12.

⇒ x³ - 1/x³ = 76.

Answer 2

QUESTION:-

$\sf\;If\;x-\frac{1}{x}=4,find\;the\;value\;of\;x^3+\frac{1}{x^3}$

EXPLANATION:-

$\sf\;\huge\;Given\;x-\frac{1}{x}=4$

Cubing both sides we get:-

$\sf\;(x-\frac{1}{x})^3=(4)^3\\\\(x)^3-3(x)^2(\frac{1}{x})+3(x)-(\frac{1}{x})^3=64\\\\\;x^3-3(x-\frac{1}{x})-\frac{1}{x^3}=64\\\\Putting\;the\;value\;of\;x-\frac{1}{x}\;in\;the\;above\;equation\;we\;get:-\\\\x^3-3(x-\frac{1}{x})-\frac{1}{x^3}=64\\\\x^3-3(4)-\frac{1}{x^3}=64\\\\x^3-12-\frac{1}{x^3}=64\\\\x^3-\frac{1}{x^3}=64+12\\\\\underline\;x^3-\frac{1}{x^3}=76$

Hence $\sf\;x^3-\frac{1}{x^3}=76$

EXTRA INFORMATION:-

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