Question

if a+b+c = 5 and ab+bc+CA =10
then prove that a^3+b^3+ c^3 = -25

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Answer 1

Step-by-step explanation:

we have

(a^3+b^3-3abc) =

(a+b+c) (a^2+b^2+c^2-ab-bc-ca)

= (a+b+c) [(a+b+c) ^2 - 3(ab+bc+ca)]

= 5×[(15)^2 - (3× 10)]

= 5 × (25-30) = 5× (-5)

= -25

Hence, (a^3+b^3+c^3 - 3abc) = -25

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Answer 2

Correct Question:-

If a + b + c = 5 and ab + bc + ca = 10 then prove that a³ + b³ + c³ - 3abc = - 25 ?

Answer:-

Given:

a + b + c = 5 -- equation (1)

ab + bc + ca = 10 -- equation (2).

We know that,

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

Substitute the value of eq - (1) here,

→ a³ + b³ + c³ - 3abc = ( a² + b² + c² - ( ab + bc + ca )(5)

Again substitute eq - (2) .

→ a³ + b³ + c³ - 3abc = ( 5 ) ( a² + b² + c² - ( 10 ) -- equation (3)

We know that,

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

→ a² + b² + c² = (5)² - 2(10)

→ a² + b² + c² = 25 - 20 = 5

Substitute (a² + b² + c² = 5) in equation (3).

→ a³ + b³ + c³ = ( 5 )( 5 - 10 )

→ a³ + b³ + c³ = 5(-5)

→ a³ + b³ + c³ = - 25

Hence, Proved.