If x^{2}+x=5,then find the value of (x+3)^{3} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $a = x+3$. We need to find the value of $a^{3} + \frac{1}{a^{3}}$.Given $x^{2} + x = 5$,we can write $x^{2} = 5 - x$.Now,consider $a = x+3$. T

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(D) Let $a = x+3$. We need to find the value of $a^{3} + \frac{1}{a^{3}}$.Given $x^{2} + x = 5$,we can write $x^{2} = 5 - x$.Now,consider $a = x+3$. Then $x = a - 3$.Substitute $x = a - 3$ into the given equation $x^{2} + x = 5$:$(a - 3)^{2} + (a - 3) = 5$$(a^{2} - 6a + 9) + a - 3 = 5$$a^{2} - 5a + 6 = 5$$a^{2} - 5a + 1 = 0$Divide the entire equation by $a$ (since $a 
eq 0$):$a - 5 + \frac{1}{a} = 0$$a + \frac{1}{a} = 5$We know that $a^{3} + \frac{1}{a^{3}} = (a + \frac{1}{a})^{3} - 3(a + \frac{1}{a})$.Substituting the value $a + \frac{1}{a} = 5$:$a^{3} + \frac{1}{a^{3}} = 5^{3} - 3(5) = 125 - 15 = 110$.

If $x^{2}+x=5$,then find the value of $(x+3)^{3}+\frac{1}{(x+3)^{3}}$.

Similar Questions

If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $ax^2 + bx + c = 0$,then:

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$
$II.$ $y^{5} - \frac{(28)^{1/2}}{\sqrt{y}} = 0$

If $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{c^{2}+a^{2}}+\frac{m-c^{2}}{a^{2}+b^{2}}=3,$ then the value of $m$ is

The number of real roots of the equation $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$ is:

The roots of the equation $ax^2 + bx + c = 0$ will be reciprocal if

Vedclass Test Series

Exam Paper Generator

Online Exam Module