Question

If a + p = 2, prove that a² + 6ap +p3 - 8 = 0

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:a + p = 2$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\: {a}^{3} + 6ap + {p}^{3} - 8 = 0$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:a + p = 2$

$\rm :\longmapsto\:a + p - 2 = 0$

We know,

$\rm :\longmapsto\:If \: a + b + c = 0 \: then \: {a}^{3} + {b}^{3} + {c}^{3} = 3abc$

Using this identity,

$\rm :\longmapsto\: {a}^{3} + {p}^{3} + {( - 2)}^{3} = 3(a)(p)( - 2)$

$\rm :\longmapsto\: {a}^{3} + {p}^{3} - 8= - 6ap$

$\rm :\longmapsto\: {a}^{3} + {p}^{3} - 8 + 6ap = 0$

$\rm :\longmapsto\: {a}^{3} + {p}^{3} + 6ap - 8 = 0$

Hence, proved

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)