Question

factorise p^3 q^3 + 64​

Answer 1

Answer:

(pq+8)(p²q²+16-4pq)

Step-by-step explanation:

p³q³+64

=(pq)³+(4)³

since (a+b)³=(a+b)(a²+b²-ab) [a=pq & b=4]

=(pq+8)(p²q²+16-4pq)

If this answer is incorrect, I apologize..

Answer 2

Answer:

(pq+4)(p^2q^2+16+4pq)

Step-by-step explanation:

by the Equation we can see there are two cubesd getting added,

so by the Identity [a^3+bb^2 +ab)^3----->  (a+b)(a^2 +]

here - p^3q^3 +64

can be written as - (pq)^3 +(4)^3

Now we can clearly see that A=Pq and B= 4

puting in the Identity we get -p(q+4)(p^2q^2+16+4pq)

Hope You Got it!