If m = -4 and n = -2,then the value of m^{3} | vedclass

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

Question

(D) Given: $m = -4$ and $n = -2$.We need to evaluate the expression: $m^{3} - 3m^{2} + 3m + 3n + 3n^{2} + n^{3}$.To simplify,we can rearrange the term

Vedclass · Accepted Answer

$-126$

Vedclass · Answer

(D) Given: $m = -4$ and $n = -2$.We need to evaluate the expression: $m^{3} - 3m^{2} + 3m + 3n + 3n^{2} + n^{3}$.To simplify,we can rearrange the terms to form perfect cubes:$(m^{3} - 3m^{2} + 3m - 1) + (n^{3} + 3n^{2} + 3n + 1) = 0$Note that we subtracted $1$ and added $1$ to keep the expression equivalent.This simplifies to: $(m - 1)^{3} + (n + 1)^{3}$.Now,substitute the values of $m$ and $n$:$(-4 - 1)^{3} + (-2 + 1)^{3}$$= (-5)^{3} + (-1)^{3}$$= -125 - 1$$= -126$.

If $m = -4$ and $n = -2$,then the value of $m^{3} - 3m^{2} + 3m + 3n + 3n^{2} + n^{3}$ is

Similar Questions

If $\lambda$ is the ratio of the roots of the quadratic equation in $x$,$3m^2x^2 + m(m - 4)x + 2 = 0$,then the least value of $m$ for which $\lambda + \frac{1}{\lambda} = 1$ is:

The coefficient of $x$ in the equation $x^2 + px + q = 0$ was taken as $17$ in place of $13$. Its roots were found to be $-2$ and $-15$. The roots of the original equation are:

Find the sum of all integral values of $a$ such that the equation $||x - 2| - |3 - x|| = 2 - a$ has a solution.

Solve the given two equations and select the correct option.
$I. 3x^2 + 5x - 2 = 0$
$II. 2y^2 - 7y + 5 = 0$

The number of integral values of $m$ for which the quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$ is always positive for all $x \in R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module