Answer

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**Hint:**Given polynomial is of degree $3$. Polynomials of degree $3$ are known as cubic polynomials. The given cubic polynomial can be converted into a quadratic polynomial by factoring out the x that is common in all the terms of the polynomial. So, the given polynomial can be converted into a cubic polynomial by the above mentioned process. Quadratic polynomials can be factored by the help of splitting the middle term method. In this method, the middle term is split into two terms in such a way that the polynomial remains unchanged.

**Complete step by step answer:**

For factorising the given cubic polynomial $y = {x^3} + 1000$ , we factor out the polynomial using algebraic identities. We know an algebraic identity of the form of sum of cubes of two terms,

$\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$

So, $y = {x^3} + 1000$

$ \Rightarrow $$y = \left( {x + 10} \right)\left( {{x^2} - 10x + 100} \right)$

Now, we have to factorise the quadratic polynomial expression thus obtained. We can use splitting the middle term method, hit and trial method or the quadratic formula to factorise the quadratic polynomial. But the obtained quadratic polynomial ${x^2} - 10x + 100$ is not factorable as the discriminant is negative. So, $y = \left( {x + 10} \right)\left( {{x^2} - 10x + 100} \right)$ is the final simplified form of the given polynomial $y = {x^3} + 1000$.

**Hence, the factored form of the cubic polynomial $y = {x^3} + 1000$ is $y = \left( {x + 10} \right)\left( {{x^2} - 10x + 100} \right)$.**

**Note:**In such questions, we are required to solve a cubic equation or factorise a cubic polynomial, we first find a root of cubic expression. Then, by factor theorem, we get a factor of the cubic polynomial and divide the cubic expression by the factor to get the remaining two roots.

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