Question

Please solve this neatly!

Answer 1

a + b + c = 5 (Given)

ab + bc + ca = 10 (Given)

Find (a² + b² + c²):

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

Sub a + b + c = 5 and ab + bc + ca = 10:

(5)² = a² + b² + c² +2(10)

a² + b² + c² = (5)² -2(10)

a² + b² + c² = 25 -20

a² + b² + c² = 5

Find (a³ + b³ + c³ - 3abc) :

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

Sub a + b + c = 5, ab + bc + ca = 10 and a² + b² + c² = 5:

a³ + b³ + c³ - 3abc = (5)(5 - 10)

a³ + b³ + c³ - 3abc = (5)(-5)

a³ + b³ + c³ - 3abc = - 25 (Proven)

Answer 2

Here's your answer !!

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It's given that,

$= > a + b + c = 5$

$= > ab + bc + ca = 10$

We know that,

$= > {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} \\ \: \: \: \: \: \: \: \: + 2(ab + bc + ca)$

By substituting the value of (a+c+b) and (ab+bc+ca) ,we get :-

$= > {(5)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(10)$

$= > 25 - 2(10) = {a}^{2} + {b}^{2} + {c}^{2}$

$= > 25 - 20 = {a}^{2} + {b}^{2} + {c}^{2}$


Therefore,

$= > {a}^{2} + {b}^{2} + {c}^{2} = 5$

Now,

We know that,

$= > {a}^{3} + {b}^{3} + {c}^{3} - 3abc= (a + b + c)( {a}^{2} + {b}^{2} \\ + {c}^{2} - ab - bc - ca )$

$= > {a}^{3} + {b}^{3} + {c}^{2} - 3abc = (a + b + c )( {a}^{2} + {b}^{2} \\ + {c}^{2} - (ab + bc + ca)$

By substituting value of (a+B+C),(ab+bc+ca)and (a^2+b^2+c^2), we get :-

$= > {a}^{3} + {b}^{3 } + {c}^{3} - 3abc =$

$= (5)(5 - 10) \\ = 5 \times( - 5) \\ = - 25$


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Hope it helps you !! :)