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**Hint:**We will first find the root of it using hit and trial method. After that factor, we will then just factorise it as we see the modifications.

**Complete step-by-step answer:**

We are given that we need to find the factorisation of $P(x) = {x^3} - 27x - 54$

We will try to find its root using the hit and trial method at first.

Let us try to put x = 1 and see what we get:

$ \Rightarrow P(1) = {1^3} - 27 - 54 \ne 0$

Now let us try to put x = -2 and see what we get:

$ \Rightarrow P( - 2) = {\left( { - 2} \right)^3} + 2 \times 27 - 54 \ne 0$

Now let us try to put x = -3 and see what we get:

$ \Rightarrow P( - 3) = {\left( { - 3} \right)^3} + 3 \times 27 - 54 = 0$

We see that clearly x = -3 is the root of the equation.

Therefore, (x + 3) is definitely a factor of the given P(x).

Now, we write the given cubic equation as follows:

$ \Rightarrow P(x) = (x + 3)f(x)$

Now, we will find the quadratic equation f(x) using division.

Let us divide P(x) by (x + 3) to get the following:-

$ \Rightarrow P(x) = (x + 3)\left( {{x^2} - 3x - 18} \right)$

Making the factors of quadratic in above expression as follows:-

$ \Rightarrow P(x) = (x + 3)\left( {{x^2} + 3x - 6x - 18} \right)$

Taking some factors common in above expression as follows:-

$ \Rightarrow P(x) = (x + 3)\left\{ {x(x + 3) - 6(x + 3)} \right\}$

We can rewrite the above expression as follows:-

$ \Rightarrow P(x) = (x + 3)(x + 3)(x - 6)$

Clubbing the like terms as follows:-

$ \Rightarrow P(x) = {(x + 3)^2}(x - 6)$

**Hence, the answer is: $P(x) = {x^3} - 27x - 54 = {(x + 3)^2}(x - 6)$**

**Note:**

The students must note that solving an equation by using factorization helps us in many aspects because we do not have to use the formula for roots of a quadratic equation which itself involves a lot of calculations and can lead to mathematical errors.

The students must also note that if they need to find the roots of equation by putting it equal to zero, we need to use a theorem for that:

It is as follows:- If a.b = 0, then either a = 0 or b = 0 or both a = b = 0.

We used this statement to get the required values of x, which is the roots of the given quadratic equation.

The students must note that an equation has as many roots as its degree. In quadratic equations, the degree is 2, so, we have 2 roots of a quadratic equation, they may be real and distinct as happened in the above question, they may be real and equal or they may be imaginary and exist in conjugates, this all depends upon the discriminant of the equation Here, in the example above, we had an equation of degree 3 so it had three roots -3 twice and 6.

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