Question

If a + b + c = 6, $a^2 + b^2 + c^2 = 30$ and $a^3 + b^3 + c^3 = 165,$ then the value of 4abc is:

• A. -1
• B. -4
• C. 1
• D. 4

Answer 1

Answer:

D. 4

Solution:

Given:

If a + b + c = 6, $a^2+b^2+c^2=30$ and $a^3+b^4+c^3=165$

Concept used:

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

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

Calculation:

⇒ $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$

⇒ 6² = 30 + 2 (ab + bc + ca)

⇒ (ab + bc + ca) = $\frac{6}{2} = 3$

⇒ $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

⇒ $165-3abc = 6\times{(30-3)}$

⇒ $165-3abc = 6\times{27}$

⇒ 3abc = 165 - 162

⇒ abc = 1

∴ $4abc = 4\times{1} = 4$