Question

Whats The coefficient of x in the expansion of (x+3)^3 ..?

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Explanation required !

Answer 1

Answer:

( x + 3 )³

⇒ x³ + (3)³ + 3 * x * 3 ( x + 3 )

⇒ x³ + 27 + 9x ( x + 3 )

⇒ x³ + 27 + 9x² + 27x

On rearranging terms:

⇒ x³ + 9x² + 27x + 27

Here, the coefficient of x is 27.

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★ Identity used:

• (a + b)³ = a³ + b³ + 3ab (a + b)

Answer 2

Answer:

27

Step-by-step explanation:

We know that,

General term of the expansion (a + b)ⁿ is:

$T_{r+1} = C(n,r)a^{n-r}b^{r}$

For general term of the expansion (x + 3)³

Putting a = x, b = 3, n = 3

$\Longrightarrow T_{r+1} = C(3,r)x^{3-r}3^{r} ------ (1)$

Now We need to find the coefficient of x.

So,

$\Longrightarrow x^{3-r} = x^1$

$\Longrightarrow 3 - r = 1$

$\Longrightarrow r = 2$

Substitute r = 2 in (1), we get

$\Longrightarrow T_{2+1} = C(3,2)x^{3-2}3^{2}$

$\Longrightarrow T_{2+1} = 3 * x^1 * 9$

$\Longrightarrow T_{2+1} = 27x$

Hence, coefficient of x = 27

Hope it helps!