Question

ifa+b+c=8and ab+bc+ac=20 find a^3+b^3+c^3-3abc

Answer 1

HELLO.......FRIEND!!

THE ANSWER IS HERE,

a+b+c= 8

Squaring on both sides.

$= > \: ({a + b + c})^{2} = {8}^{2}$

$= > \: {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 64$

$= > \: {a}^{2} + {b}^{2} + {c}^{2} + 2(20) = 64$

$= > \: {a}^{2} + { b}^{2} + {c}^{2} = 64 - 40 = 24$

We know the formula,

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

$= > \: 8(24 - 20)$
$= > \: 8 \times 4$

$= > \: 32$

$so \: {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 32$


:-)Hope it helps u.

Answer 2

Hi Friend !!

THE ANSWER IS HERE,

=======================
a+b+c= 8

Squaring on both sides.

(a+b+c)² = 8²

a²+b²+c²+2(ab+bc+ca) = 64

a²+b²+c²+2(20) = 64

a²+b²+c² = 64-40

a²+b²+c² = 24


We know the formula,

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

= 8(24-20)

= 8(4)

= 32



Hope it helps u