Prove that the expansion of $(1 - x^{3})^{n}$ may be put into the form $(1 - x)^{3 n} + 3 n x (1 - x)^{3 n - 2} + \frac{3 n (3 n - 3)}{1.2} x^{2} (1 - x)^{3 n - 4} + . . .$

Open in App

Solution

we know that
$1 - x^{3} = (1 - x)^{3} + 3 x (1 - x)$

$(1 - x^{3})^{n} = [(1 - x)^{3} + 3 x (1 - x)]^{n}$

taking $(1 - x)^{3}$ common in RHS

$= (1 - x)^{3 n} . [1 + 3 x (1 - x)^{- 2}]^{n}$

now expand it

$= (1 - x)^{3 n} . [1 + 3 n x (1 - x)^{- 2} + \frac{(n) (n - 1) .9 x^{2} . (1 - x)^{- 4}}{1.2} . . . . . . . . .]$

now solving on multiplying with outer term

$= (1 - x)^{3 n} + (1 - x)^{3 n} .3 n x (1 - x)^{- 2} + (1 - x)^{3 n} . \frac{(3 n) (3 n - 3) . x^{2} . (1 - x)^{- 4}}{1.2} . . . . . . . . .$

$= (1 - x)^{3 n} + 3 n x (1 - x)^{3 n - 2} + . \frac{(3 n) (3 n - 3) . x^{2} . (1 - x)^{3 n - 4}}{1.2} . . . . . . . . .$

$∴ (1 - x^{3})^{n} = (1 - x)^{3 n} + 3 n x (1 - x)^{3 n - 2} + . \frac{(3 n) (3 n - 3) . x^{2} . (1 - x)^{3 n - 4}}{1.2} . . . . . . . . .$

Hence proved

Suggest Corrections

Similar questions

{(x^{3 n + 1} . x^{3 n - 1})}^{2}

{(1 - x^{3})}^{n} = \sum_{r = 0}^{n} a_{r} x^{r} {(1 - x)}^{3 n - 2 r}

a_{r}

n \in N

x^{4}

{(1 + x + x^{2} + x^{3})}^{n}

^{n} C_{4} +^{n} C_{2} +^{n} C_{1} \cdot^{n} C_{2}

x^{4}

{(1 + x + x^{2} + x^{3})}^{n}

(n + 1) t h

{(2 x - \frac{1}{x})}^{3 n}

Join BYJU'S Learning Program