Answer:
a + b + c = m
=> (a + b+ c)³= m³
⇔m³ = a³ + b³ +c³ + 3(a + b)(a + c)(c + b)
=> m³ + 2p³ = 3a³ + 3b³ + c³ + 3(a + b)(a + c)(b + c)
with c = 0 => m³ + 2p³ = 3a³ + 3b³ + 3ab(a + b)
have:
3mn = 3(a + b + c)(a² + b² + c²)
with c = 0
=> 3mn = 3(a + b)(a² + b²) = 3[a³ + ab² + ba² + b³]
= 3a³ + 3b³ + 3(a²b + ab²)
= 3a³ + 3b³ + 3ab(a + b)
= m³ + 2p³
=> 3mn = m³ + 2p³
Step-by-step explanation:
