Question

a , b , c are three real numbers such that a + b + c = 7 , a 2 + b 2 + c 2 = 35 and a 3 + b 3 + c 3 = 151 . Find the value of abc.

Answer 1

Hi ,

a + b + c = 7 -------( 1 )

a ^2 + b^2 + c^2 = 35 ------( 2 )

a^3 + b^3 + c ^3 = 151 -----( 3 )

abc = ?

Do the square of equation ( 1 )

( a + b + c )^2 = 7^2

a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 49

(a^2 + b^2 + c^2) +2 (ab + bc + ca ) = 49

35 + 2 ( ab + bc + ca ) = 49 [ From ( 2 ) ]

2( ab + bc + ca ) = 49 - 35

ab + bc + ca = 14 /2

ab + bc + ca = 7 --------------------------( 4 )

We know the identity,
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a^3 + b^3 + c^3
= ( a + b + c ) ( a^2 + b^2 + c^2 - ab-bc-ca ) + 3abc
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151 = 7 × [ 35 - ( ab + bc + ca ) ] + 3 abc [ from (1) ,(2) ,(3)]

151 = 7 [ 35 - 7 ] + 3abc [ from ( 4 ) ]

151 = 7 × 28 + 3abc

151 = 196 + 3abc

151 - 196 = 3abc

- 45 = 3abc

( -45 ) / 3 = abc

-15 = abc

Therefore ,

abc = -15

I hope this helps you.

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Answer 2

Step-by-step explanation:

abc=-15...........

thank You bye