Question

Factorise :
p9 - q3 + 1 + 3pq​

Answer 1

Step-by-step explanation:

hope you will understand

Answer 2

Answer:

$3pq(1-p^{2})+(p^{3}-q+1)(p^{6}+q^{2}+1+p^{3}q+q-p^{3})$

Step-by-step explanation:

In the question,

The equation $p^{9}-q^{3}+1+3pq$

So,

As we know from the Property of Sum of the Cubes is,

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

So,

Using the same property we can simply write the term in the expanded form as,

$((p^{3}))^{3}+(-q)^{3}+(1)^{3}=3(p^{3}.(-q)(1))+(p^{3}-q+1)(p^{6}+q^{2}+1+p^{3}q+q-p^{3})\\p^{9}-q^{3}+1=-3p^{3}q+(p^{3}-q+1)(p^{6}+q^{2}+1+p^{3}q+q-p^{3})$

Therefore, on putting this in the given equation we get the factorised form of the equation as,

$p^{9}-q^{3}+1+3pq=3pq-3p^{3}q+(p^{3}-q+1)(p^{6}+q^{2}+1+p^{3}q+q-p^{3})\\p^{9}-q^{3}+1+3pq=3pq(1-p^{2})+(p^{3}-q+1)(p^{6}+q^{2}+1+p^{3}q+q-p^{3})$

Therefore, the Final factorised equation is given by the term,

$3pq(1-p^{2})+(p^{3}-q+1)(p^{6}+q^{2}+1+p^{3}q+q-p^{3})$