Question

If p + q = 1 + pq, prove that p3 + q3 = 1 + p3q3

Answer 1

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

$\rm :\longmapsto\:p + q = 1 + pq$

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

$\rm :\longmapsto\: {p}^{3} + {q}^{3} = 1 + {p}^{3} {q}^{3}$

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

We know,

$\green{ \boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}}$

Given that,

$\rm :\longmapsto\:p + q = 1 + pq$

On cubing both sides, we get

$\rm :\longmapsto\: {(p + q)}^{3} = {(1 + qp)}^{3}$

$\rm :\longmapsto\: {p}^{3} + {q}^{3} + 3pq(p + q) = {1}^{3} + {(pq)}^{3} + 3pq(1 + pq)$

$\rm :\longmapsto\: {p}^{3} + {q}^{3} + 3pq(p + q) = {1}^{3} + {(pq)}^{3} + 3pq(p + q)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: p + q = 1 + pq\bigg \}}$

$\rm :\longmapsto\: {p}^{3} + {q}^{3} = 1 + {p}^{3} {q}^{3}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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)