Fractorise:
`16(2p-3q)^(2) -4 (2p - 3q)`

Fractorise: 16p^(3) -4p

Let's resolve into factors: (p^2 - 3q^2)^2 - 16 (p^2 - 3q^2) + 63

Resolve into factors : (4p-3q)^3+(3q-2)^3+(2-4p)^3

Find the product : (p^(2) + q^(2))(2p - 3q + 4r)

Find each of the following products : (3p^(2) + q^(2)) xx (2p^(2) - 3q^(2))

Find the each of the following products: (3p^(2) + q^(2)) (2p^(2) - 3q^(2))

Show that : (4p q+3q)^2-(4q p-3q)^2=48 p q^2 (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

If p and q are non-collinear unit vectors and | p+q| = sqrt(3) , then (2p - 3q) . (3p + q) is equal to

Sum the following infinite series (p-q) (p+q) + (1)/(2!) (p-q)(p+q) (p^(2) + q^(2))+(1)/(3!) (p-q) (p+q) (p^(4)+q^(4)+p^(2) q^(2)) + ...oo