Question

If a+b+c=15 and a2+b2+c2=83,find the value of a3+b3+c3-3abc.

Answer 1

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Given:-)

→ a + b + c = 15.

→ a² + b² + c² = 83.



To Find:-)

→ a³ + b³ + c³ - 3abc.


Solution:-)

▶We have,

=> a + b + c = 15.

[ Squaring both side ].

=> ( a + b + c )² = ( 15 )².

=> a² + b² + c² + 2( ab + bc + ca ) = 225.

=> 83 + 2( ab + bc + ca ) = 225.

=> 2( ab + bc + ca ) = 225 - 83.

$=> ab + bc + ca = \frac{142}{2} .$

=> ab + bc + ca = 71.


▶ Now,


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


= ( 15 ) ( 83 - ( 71 ) ).

= 15 × 12.

= 180.



✔✔ Hence, it is solved ✅✅.

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Answer 2

a + b +c = 15
And a2 + b2 + c2 = 83
Now we know that
( a + b + c) 2 = a2+ b2 +c2 + 2ab + 2bc + 2ac
225 -83 /2 = ab + bc + ca
142/2=ab+ bc + ac
71= ab + bc + ac
according to question
a3+ b3+ c3-3abc = ( a + b + c) (a2+ b2+ c2-ab - bc - ca)
= (15 )(83 - 71)
= 15*12
= 180


Hey dude may it helps ugh..