Question

If a+b+c=8 and ab+bc+ca=25 then find the value of a3+b3+c3

-3abc.​

Answer 1

Answer:

Step-by-step explanation:

$\large{\bold{a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}}\\\\\large{\bold{a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca-3(ab+bc+ca))}}\\\\\large{\bold{a^3+b^3+c^3-3abc=(a+b+c)((a+b+c)^2+3(ab+bc+ca))}}\\\\\large{\bold{a^3+b^3+c^3-3abc=(8)((8)^2+3(25))}}\\\\\large{\bold{a^3+b^3+c^3-3abc=(8)(64-75)}}\\\\\large{\bold{a^3+b^3+c^3-3abc=-88}}$

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Answer 2

a + b + c = 8

ab + bc + ca = 25

_______ [GIVEN]

a³ + b³ + c³ - 3abc

_______ [FIND]

Solution:

=> a + b + c = 8

• Do squaring on both sides.

=> (a + b + c)² = (8)²

=> a² + b² + c² + 2ab + 2bc + 2ca = 64

=> a² + b² + c² + 2(ab + bc + ca) = 64

=> a² + b² + c² + 2(25) = 64

=> a² + b² + c² + 50 = 64

=> a² + b² + c² = 64 - 50

=> a² + b² + c² = 14

______________________________

We have to find a³ + b³ + c³ - 3abc

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

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

=> a³ + b³ + c³ - 3abc = (8) [14 - 25]

=> a³ + b³ + c³ - 3abc = 8 (-11)

______________________________

a³ + b³ + c³ - 3abc = - 88

_____________ [$\bold{ANSWER}]$