Question : If (a + b + c) = 12 and (ab + bc + ca) = 47, find the value of (a3 + b3 + c3 – 3abc).
Option 1: 24
Option 2: 36
Option 3: 48
Option 4: 42
Correct Answer: 36
Solution :
Given: (a + b + c) = 12 and (ab + bc + ca) = 47.
On squaring both sides of the equation, (a + b + c) = 12, we get,
(a + b + c)
2
= 12
2
⇒ a
2
+ b
2
+ c
2
+ 2(ab + bc + ca) = 144
⇒ a
2
+ b
2
+ c
2
+ 2(47) = 144
⇒ a
2
+ b
2
+ c
2
= 144 – 94
⇒ a
2
+ b
2
+ c
2
= 50
The value of (a
3
+ b
3
+ c
3
– 3abc) is,
a
3
+ b
3
+ c
3
– 3abc = (a + b + c)(a
2
+ b
2
+ c
2
– (ab + bc + ca))
⇒ a
3
+ b
3
+ c
3
– 3abc = 12(50 – 47)
⇒ a
3
+ b
3
+ c
3
– 3abc = 12 × 3
$\therefore$ a
3
+ b
3
+ c
3
– 3abc = 36
Hence, the correct answer is 36.
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