If $a + b + c = 8$ and $a b + b c + c a = 20$ , find the value of $a^{3} + b^{3} + c^{3} - 3 a b c$ .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 32
Since ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
$a + b + c = 8 and a b + b c + c a = 20$
${(8)}^{2} = a^{2} + b^{2} + c^{2} + 2 \times 20$
$64 = a^{2} + b^{2} + c^{2} + 40$
$a^{2} + b^{2} + c^{2} = 24$
We now use the following identity:
$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - (a b + b c + c a))$
$a^{3} + b^{3} + c^{3} - 3 a b c = 8 \times (24 - 20) = 4 \times 8 = 32$
Thus, $a^{3} + b^{3} + c^{3} - 3 a b c = 32$

Suggest Corrections

Similar questions

a + b + c = 8

a b + b c + c a = 20

a^{3} + b^{3} + c^{3} - 3 a b c

a + b + c = 9

ab + bc + ca = 26,

a^{3} + b^{3} + c^{3} - 3abc

If a+b+c=8 and ab+ bc+ca=20, find the value for a³+b³+c³-3abc

the options are :-

please solve this question with explanation

a + b + c = 11

a b + b c + c a = 20

a^{3} + b^{3} + c^{3} - 3 a b c

Join BYJU'S Learning Program