Question

if p=2-a then prove that a³+6ap+p³-8=0

Answer 1

Solution:

We have,

p = 2 - a

➲ a + p - 2 = 0

➲ x + y + z = 0

where a = x , p = y and (-2) = z

➲ x³ + y³ + z³ = 3xyz

where [ x+y+z = 0 => x³ + y³ + z³ = 3xyz]

➲ a³ + p³ + (-2)³

➲ 3 × a × p × (-2)

Then,

➲ a³ + 6ap + p³ - 8 = 0

Hence,

Proved that a³ + 6ap + p³ - 8 = 0

Extra Dose:

• (a + b)³ = a³ + b³ + 3ab (a+b)
• (a - b)³ = a³ - b³ - 3ab (a -b)

Thanks: