Question

If a + b + c = 10 and ab + bc + ac = 31, Find the value of
a^3 +b^3 + c^3 – 3abc​

Answer 1

Answer is 70.

${a}^{3} + {b}^{ 3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)$We have all the values except a²+b²+c². To find it, we use the identity

${(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca$

Putting the values we get

${(10)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2 \times 31$

$100 = {a}^{2} + {b}^{2} + {c}^{2} + 62$

${a}^{2} + {b}^{2} + {c}^{2} = 100 - 62 = 38$

Now putting in the values in the first equation we get

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

${a}^{3} + {b}^{ 3} + {c}^{3} - 3abc = 10 \times (38 - 31) = 10 \times 7 = 70$