Question

If 1/6x²-7x+2 = A/3x-2 +B/2x-1, then A³+B³=​

Answer 1

Answer:

The value of  $A^3+B^3$ is 19

Step-by-step explanation:

Given:

$\frac{1}{6x^2-7x+2}=\frac{A}{3x-2}+\frac{B}{2x-1}$

​

$\implies\frac{1}{(3x-2)(2x-1)}=\frac{A(2x-1)+B(3x-2)}{(3x-2)(2x-1)}$

$\implies\,1=A(2x-1)+B(3x-2)$

$put\,x=\frac{1}{2}$

$1=A(2(\frac{1}{2})-1)+B(3(\frac{1}{2})-2)$

$1=B(\frac{3}{2}-2)$

$1=B(\frac{-1}{2})$

$\implies\boxed{\bf\,B=-2}$

$put\,x=\frac{2}{3}$

$1=A(2(\frac{2}{3})-1)+B(3(\frac{2}{3})-2)$

$1=A(\frac{1}{3})$

$\implies\boxed{\bf\,A=3}$

Now

$A^3+B^3$

$=(3)^3+(-2)^3$

$=27-8$

$=19$

Answer 2

The value of A³+B³ is 19

Step-by-step explanation:

$\dfrac{1}{6x^2-7x+2}=\dfrac{A}{3x-2}+\dfrac{B}{2x-1}$

Using partial fraction,

$\dfrac{1}{6x^2-7x+2}=\dfrac{A(2x-1)+B(3x-2)}{(3x-2)(2x-1)}$

$\dfrac{1}{6x^2-7x+2}=\dfrac{2Ax-A+3Bx-2B}{6x^2-7x+2}$

Compare the coefficient of numerator and make equation.

$1=(2A+3B)x-A-2B$

2A+3B=0 and

-A-2B=1  multiply by 2

-2A-4B=2

-B = 2

B = -2

A = 3

Now calculate the value of A³+B³

$\Rightarrow 3^3+(-2)^3$

$\Rightarrow 27-8$

$\Rightarrow 19$

Hence, the value of A³+B³ is 19

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