Question

if a + b + C is equal to 6 a square + b square + c square is equal to 14 and a cube plus b cube plus c cube is equal to 36 then prove that ABC is equal to 6

Answer 1

hey mate
here's the solution

Answer 2

Solution :

It is given that ,

a + b + c = 6 ----( 1 )

a² + b² + c² = 14 ---( 2 )

a³ + b³ + c³ = 36 ---( 3 )

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We know the algebraic identities,

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

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

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Here ,

Do the square of equation ( 1 ),

( a+b+c)² = 6²

=> a²+b²+c² + 2(ab+bc+ca) = 36

=> 14 + 2(ab+bc+ca ) = 36

=> ab+bc+ca = 22/2

=> ab+bc+ca = 11 ---( 4 )

Now ,

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

=>36 -3abc = 6[14 - 11 ]

=> 36 - 3abc = 6 × 3

=> 36 - 18 = 3abc

=> 18 = 3abc

=> 18/3 = abc

=> 6 = abc


Therefore ,

abc = 6

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