$8 x^{3} + 12 x^{2} + 6 x + 1 = (A x + B)^{3}$ then $A + B =$ ......

1

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Solution

The correct option is C $3$
Given that,
$8 x^{3} + 12 x^{2} + 6 x + 1 = (A x + B)^{3}$

$⟹ (2 x)^{3} + 12 x^{2} + 6 x + (1)^{3} = (A x + B)^{3}$

$⟹ (2 x)^{3} + 3 (2 x)^{2} (1) + 3 (2 x) (1)^{2} + (1)^{3} = (A x + B)^{3}$

$⟹ (2 x + 1)^{3} = (A x + B)^{3}$

By comparing, we get
$A = 2, B = 1$

$∴ A + B = 2 + 1 = 3$

Option B is correct.

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t a n^{- 1} [\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}] - t a n^{- 1} [\frac{4 x}{1 - 4 x^{2}}] = t a n^{- 1} 2 x, | 2 x | < \frac{1}{\sqrt{3}}

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

x : {tan}^{- 1} (x - 1) + {tan}^{- 1} x + {tan}^{- 1} (x + 1) = {tan}^{- 1} 3 x

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

8 x^{3} + \frac{1}{8 x^{3}} + \frac{3}{2 x} + 6 x

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