Question

Write the given cube in expanded form: $$\left( 2p-5\right)^{3} $$

Answer 1

Hint: In this question it is given that we have to find the expanded form of $$\left( 2p-5\right)^{3} $$. So to find the expanded form we need to know the cubic identity, i.e, if any cube of a binomial is given $$\left( a-b\right)^{3} $$ then it can be expanded as,
$$\left( a-b\right)^{3} =a^{3}-3a^{2}b+3ab^{2}-b^{3}$$.....(1)

Complete step-by-step answer:
Given, $$\left( 2p-5\right)^{3} $$
Now comparing the above cube with $$\left( a-b\right)^{3} $$, we can write,
a = 2p and b = 5
Now putting the values of a and b in equation (1), we get,
$$\left( a-b\right)^{3} =a^{3}-3a^{2}b+3ab^{2}-b^{3}$$
$$\Rightarrow \left( 2p-5\right)^{3} =\left( 2p\right)^{3} -3\times \left( 2p\right)^{2} \times 5+3\times \left( 2p\right) \times 5^{2}+5^{3}$$
Now as we know that, $$\left( ab\right)^{n} =a^{n}\times b^{n}$$,
So by the formula we can write the above equation as,
$$\left( 2p-5\right)^{3} =2^{3}\times p^{3}-3\times 2^{2}\times p^{2}\times 5+3\times 2\times p\times 5^{2}+5^{3}$$
$$\Rightarrow \left( 2p-5\right)^{3} =8p^{3}-(3\times 4\times p^{2}\times 5)+(3\times 2\times p\times 25)+125$$ [since, $5^{3} =125$]
$$\Rightarrow \left( 2p-5\right)^{3} =8p^{3}-60p^{2}+150p+125$$
Hence the expanded form of $$\left( 2p-5\right)^{3} $$ is $$8p^{3}-60p^{2}+150p+125$$


Note: If you are asked to find the expanded form of cube of a term by not using the identity, then you can also find the cube of a binomials (2p-5) i.e, $$(2p-5)^{3}$$ just by multiplying the term three times,
$$\left( 2p-5\right)^{3} =\left( 2p-5\right) \left( 2p-5\right) \left( 2p-5\right) $$
Where first multiply the first two binomials and after that multiply the third binomial with the resultant.