Question 106
If $$a + b + c = 19, ab + bc + ca = 120$$, then what is the value of $$a^3 + b^3 + c^3 - 3abc$$?
Solution
$$(a + b + c)^2 = a^2 +b^2 + c^2 + 2ab + 2bc + 2ca$$
$$ 19^2 = a^2 +b^2 + c^2 + 2(120)$$
$$a^2 +b^2 + c^2 = 361 - 240 = 121$$
$$a^3 + b^3 + c^3 - 3abc$$
=$$ (a + b + c)(a^2 + b^2 + c^2 -ab -bc - ca)$$
= 19(121$$ \times -120) = 19$$
Get AI Help
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
