What is p-2(p3-p4)+p-1(pq+p2a)?

Answer 1

Given p=2−a

to prove a

3

+6ap+p

3

−8=0...(1)

Now put the value of P in eq (1)

a

3

+6a(2−a)+(2−a)

3

−8=0

⇒a

3

+12a−6a

2

+8−a

3

−12a+6a

2

−8=0

[∴(a−b)

3

=a

3

−b

3

−3a

2

b+3ab

2

]

⇒0=0 (By eliminating equal terms we

have LHS = RHS)

Answer 2

the answer is 2p4-2p3-ap2+p-pq

4

​

−2p

3

​

−ap

2

​

+2p−pq