plzzzz solve urgent
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Given (a + b + c) = 12
Squaring on both the sides, we get
(a + b + c)2 = 12²
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 144
⇒ 90 + 2(ab + bc + ca) = 144
⇒ 2(ab + bc + ca) = 144 – 90 = 54
⇒ ab + bc + ca = 27 → (1)
Recall that a3+b3+c3-3abc = (a + b + c)( a2 + b2 + c2 – ab – bc – ca)
= 12(90– 27)
= 12(63) = 756
Squaring on both the sides, we get
(a + b + c)2 = 12²
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 144
⇒ 90 + 2(ab + bc + ca) = 144
⇒ 2(ab + bc + ca) = 144 – 90 = 54
⇒ ab + bc + ca = 27 → (1)
Recall that a3+b3+c3-3abc = (a + b + c)( a2 + b2 + c2 – ab – bc – ca)
= 12(90– 27)
= 12(63) = 756
Pari0210:
sorry.... in the last 2nd step it will be 12(63) that is equal to 756
sorry for the mistake
You can edit it , sis
yeb bro, did ... thnx
yes*
Ur wlcm sis !
:-)
thanks for marking me as brainliest , bro
Answered by
1
Given,
⇒ ( a + b + c ) = 12 ------- ( 1 )
⇒ ( a² + b² + c² ) = 90 ------- ( 2 )
We have to find the value of ( a³ + b³ + c³ - 3abc ).
Using identity,
⇒ a³ + b³ + c³ - 3abc = ( a + b + c ) ( a² + b² + c² - ab - bc - ca )
Here,we can find that we have the values of ( a + b + c ) and ( a² + b² + c² ) but to find the value of ( a³ + b³ + c³ - 3abc ), we need one more value i.e. ( -ab - bc - ca ).
We can find this value by using eq.( 1 ) and eq.( 2 ).
Let's do it !
We have,
⇒ ( a + b + c ) = 12 as equation ( 1 ).
Squaring both sides,
⇒ ( a + b + c )² = 12²
Using identity,
⇒ ( a + b + c )² = a² + b² +c² + 2ab + 2bc + 2ca
⇒a² + b² + c² + 2ab + 2bc + 2ca = 144
Substitute the value of ( 2 ),
⇒90 + 2ab + 2bc + 2ca = 144
⇒ 2ab + 2bc + 2ca = 144 - 90
Taking out 2 as common in L.H.S,
⇒ 2( ab + bc + ca ) = 54
⇒ ( ab + bc + ca ) = 54 ÷ 2
⇒ ( ab + bc + ca ) = 27
Changing their sides,
⇒ -27 = -ab - bc - ca -------- ( 3 )
Now, we have the values of all necessary terms to find our final solution.
= a³ + b³ + c³ - 3abc
= ( a + b + c ) ( a² + b² + c² - ab - bc - ca )
Substitute the value of ( 1 ) , ( 2 ) and ( 3 ),
= ( 12 ) ( 90 - 27 )
= 12 × 63
= 756
The required answer is 756.
⇒ ( a + b + c ) = 12 ------- ( 1 )
⇒ ( a² + b² + c² ) = 90 ------- ( 2 )
We have to find the value of ( a³ + b³ + c³ - 3abc ).
Using identity,
⇒ a³ + b³ + c³ - 3abc = ( a + b + c ) ( a² + b² + c² - ab - bc - ca )
Here,we can find that we have the values of ( a + b + c ) and ( a² + b² + c² ) but to find the value of ( a³ + b³ + c³ - 3abc ), we need one more value i.e. ( -ab - bc - ca ).
We can find this value by using eq.( 1 ) and eq.( 2 ).
Let's do it !
We have,
⇒ ( a + b + c ) = 12 as equation ( 1 ).
Squaring both sides,
⇒ ( a + b + c )² = 12²
Using identity,
⇒ ( a + b + c )² = a² + b² +c² + 2ab + 2bc + 2ca
⇒a² + b² + c² + 2ab + 2bc + 2ca = 144
Substitute the value of ( 2 ),
⇒90 + 2ab + 2bc + 2ca = 144
⇒ 2ab + 2bc + 2ca = 144 - 90
Taking out 2 as common in L.H.S,
⇒ 2( ab + bc + ca ) = 54
⇒ ( ab + bc + ca ) = 54 ÷ 2
⇒ ( ab + bc + ca ) = 27
Changing their sides,
⇒ -27 = -ab - bc - ca -------- ( 3 )
Now, we have the values of all necessary terms to find our final solution.
= a³ + b³ + c³ - 3abc
= ( a + b + c ) ( a² + b² + c² - ab - bc - ca )
Substitute the value of ( 1 ) , ( 2 ) and ( 3 ),
= ( 12 ) ( 90 - 27 )
= 12 × 63
= 756
The required answer is 756.
:-)
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