If (x-a)(x-b)=1 and a-b+5=0, then the value o | vedclass

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

Question

(D) Given $(x-a)(x-b)=1$,we can write $(x-b) = \frac{1}{x-a}$.We are given $a-b+5=0$,which implies $a-b = -5$,or $b-a = 5$.We need to find the value o

Vedclass · Accepted Answer

$140$

Vedclass · Answer

(D) Given $(x-a)(x-b)=1$,we can write $(x-b) = \frac{1}{x-a}$.We are given $a-b+5=0$,which implies $a-b = -5$,or $b-a = 5$.We need to find the value of $(x-a)^{3}-\frac{1}{(x-a)^{3}}$.Substitute $\frac{1}{x-a} = x-b$ into the expression:$(x-a)^{3} - (x-b)^{3}$.Using the identity $A^3 - B^3 = (A-B)(A^2 + AB + B^2)$ or simply substituting $x-b = x-a + (a-b)$:Let $u = x-a$. Then $x-b = u + (a-b) = u - 5$.So,the expression becomes $u^3 - (u-5)^3$.Expanding this: $u^3 - (u^3 - 15u^2 + 75u - 125) = 15u^2 - 75u + 125$.Alternatively,using the given $(x-a)(x-b)=1$:$u(u-5) = 1 \implies u^2 - 5u = 1$.We want $u^3 - (u-5)^3 = u^3 - (u^3 - 3u^2(5) + 3u(25) - 125) = 15u^2 - 75u + 125$.$= 15(u^2 - 5u) + 125$.Substituting $u^2 - 5u = 1$:$= 15(1) + 125 = 15 + 125 = 140$.

If $(x-a)(x-b)=1$ and $a-b+5=0,$ then the value of $(x-a)^{3}-\frac{1}{(x-a)^{3}}$ is:

Similar Questions

The value of $\left(2-\frac{1}{3}\right)\left(2-\frac{3}{5}\right)\left(2-\frac{5}{7}\right) \ldots \left(2-\frac{999}{1001}\right)$ is

The value of $\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$ is

The value of $\frac{2.75 \times 2.75 \times 2.75 - 2.25 \times 2.25 \times 2.25}{2.75 \times 2.75 + 2.75 \times 2.25 + 2.25 \times 2.25}$ is:

When simplified,the value of the product $\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots\left(1-\frac{1}{n}\right)$ equals

$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module