Question

Solve the following equation for $(x\ge 0)$

$\frac{(3x-4{)}^3-(x+1{)}^3}{(3x-4{)}^3+(x+1{)}^3}=\frac{61}{189}$

• A. $\left\{3\right\}$
• B. $\left\{5\right\}$
• C. $\left\{8\right\}$
• D. $\left\{13\right\}$

Answer 1

The correct option is A $\left\{3\right\}$

$\frac{(3x-4{)}^3-(x+1{)}^3}{(3x-4{)}^3+(x+1{)}^3}=\frac{61}{189}$
Applying Compounding and Dividendo ,we get,
$=>\frac{(3x-4{)}^3-(x+1{)}^3+(3x+4{)}^3+(x+1{)}^3}{(3x-4{)}^3-(x+1{)}^3+(3x+4{)}^3+(x+1{)}^3}=\frac{61+189}{61-189}$
$=>\frac{2(3x-4{)}^3}{-2(x+1{)}^3}=\frac{200}{-128}$
$=\frac{125}{64}$
$=>\frac{(3x-4{)}^3}{(x+1{)}^3}=\frac{125}{64}$
$=>\frac{(3x-4{)}^3}{(x+1{)}^3}=\frac{5^3}{4^3}$
$\therefore 3x-4=5$ $=>x+1=4$
$=>x=\frac{9}{3}$ $=>x=3$
$=>x=3$
$\therefore x=3$