Question

a + b + c = m, a squar + b square + c square = n and a + b + c
(A) What is the value of ab + bc + ca?
(B) If a = 0, m = 3 and n 5 then what is the value ofb3 + c?
(C) If c = 0, show that m2 + 2p2 = 3 mn.​

Answer 1

Correct Question:

If a + b + c = m, a² + b² + c² = n and a³ + b³ + c³ = p³ then,

(A) What is the value of ab + bc + ca?

(B) If a = 0, m = 3 and n = 5 then what is the value of b³ + c³?

(C) If c = 0, show that m³ + 2p³ = 3 mn.

Answer:

Here we have direct values of a + b + c = m, a² + b² + c² = n & a³ + b³ + c³ = p³ and are asked to find out the value of ab + bc + ca; b³ + c³ if value of a is 0, m is 3 & c is 5 and have to show that m³ + 2p³ = 3mn if c = 0.

We will have to use the identities (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) and a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) to solve this question.

Let's start!

(A) Used identity: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

→ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Just substitute the values of a + b + c, a² + b² + c² as m and n (given) respectively to reach to our destination.

→ m² = n + 2(ab + bc + ca)

→ m² - n = 2(ab + bc + ca)

Finally divide with 2 on both sides,

→ (m² - n)/2 = 2(ab + bc + ca)/2

→ (m² - n)/2 = ab + bc + ca

Hence, the value of ab + bc + ca is (m² - n)/2.

(B) We know: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca),

Substitute the value of a = 0 in a³ + b³ + c³ - 3ac and a + b + c & a² + b² + c² as m & n (as given in question)

→ (0)² + b³ + c³ - 3(0)bc = (m)(n - ab - bc - ca)

→ 0 + b³ + c³ - 0 = m(n - ab - bc - ca)

→ b³ + c³ = m[n - (ab + bc + ca)]

From above calculations we have value of ab + bc + ca is (m² - n)/2. Using it we get,

→ b³ + c³ = m(3n - m²)/2

Furthermore we have value of m and n as 3 and 5 respectively. So,

→ b³ + c³ = 3(3*5 - 3^2)/2

→ b³ + c³ = 3/2 × (15 - 9)

→ b³ + c³ = 3/2 × 6

→ b³ + c³ = 9

Hence, the value of b³ + c³ is 9.

(C) a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Simply substitute the value of a³ + b³ + c³ = p³, a + b + c = m, a² + b² + c² = n and ab + bc + ca = (m² - n)/2.

→ p³ - 3ab(0) = m(2n - m² + n)/2 [ c = 0 given ]

→ p³ = m(3n - m²)/2

→ 2p³ = 3mn - m³

→ 2p³ + m³ = 3mn

Hence, proved.