Question

If a + b + c = 11 and ab + bc + ca = 17, then what is the value of a3 + b3 + c3 – 3abc?

A) 121 B) 168 C) 300 D) 770

Answer 1

Step-by-step explanation:

Given -

• a + b + c = 11
• ab + bc + ca = 17

→ -(-ab - bc - ca) = 17

→ -ab - bc - ca = -17

To Find -

• Value of a³ + b³ + c³ - 3abc

As we know that :-

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

Now,

a + b + c = 11

Squaring both sides :-

→ (a + b + c)² = (11)²

→ a² + b² + c² + 2(ab + bc + ca) = 121

→ a² + b² + c² + 2(17) = 121

→ a² + b² + c² = 121 - 34

→ a² + b² + c² = 87

Now,

→ a³ + b³ + c³ - 3abc = (11)(87 - 17)

→ a³ + b³ + c³ - 3abc = 11 × 70

→ a³ + b³ + c³ - 3abc = 770

Hence,

Option 3 is correct.

Answer 2

$\large{\mathbb{\underline{\underline{ANSWER}}}}$

GIVEN:-

• a+b+c=11 and ab+bc+ca=17

TO FIND:-

$a^3+b^3+c^3-3abc$

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

• How To solve?

• We have to find the value of those which are not given

• We have to find the value of $a^2+b^2+c^2$

• For this we have to use another identity i.e. $(a+b+c)^2$

Now,

$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$

$11^2=a^2+b^2+c^2+2(17)$

$121-34=a^2+b^2+c^2$

$a^2+b^2+c^2=87$

Atq.

Put the given values

$\implies{a^3+^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}$

$\implies{a^3+^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ac)}$

$\implies{(11)(87-17)}$

$\implies{(11)(70)}$

$770$

Hence,the answer is D-770