Question

solve this question.​

Answer 1

$\underline{\boxed{\text{GiVen:}}}$

• a + b + c = 14
• a² + b² + c² = 60

$\underline{\boxed{\text{To Find:}}}$

• The value of a³ + b³ +c³ -3abc

$\underline{\boxed{\text{ SoLution:}}}$

We have,

a +b +c = 14

Squaring both sides

(a + b + c)² = (14)²

We know that,

(a +b +c)² = (a² + b² + c² +2ab +2bc +2ca)

(a +b +c)² = [ a² +b² +c² +2(ab +bc +ca) ]

So,

(a + b + c)² = (14)²

[ a² +b² +c² +2(ab +bc +ca) ] = 196

Putting the value

60 +2(ab + bc + ca) = 196

2(ab +bc +ca) = 196 -60

2(ab +bc +ca) = 136

ab + bc + ca = 136/2

ab + bc + ca = 68

Now,

We also know that

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

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

Substituting values

a³ +b³ +c³ -3abc = (14) [60 - (68)]

a³ +b³ +c³ -3abc = 14 (60 -68)

a³ +b³ +c³ -3abc = 14 (-8)

$\boxed{\green{{a}^{3} +{b}^{3} +{c}^{3} -3abc = -112}}$