Question

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

Answer 1

Given that,

a + b + c = 5


Let's sqaure both sides

=> (a + b + c)² = (5)²

=> a² + b² + c² + 2(ab + bc + ca) = 25

[Using the identity]

=> a² + b² + c² + 2(10) = 25

[Since it's given that ab + bc + ca = 10)

=> a² + b² + c² + 20 = 25

=> a² + b² + c² = 25 - 20

=> a² + b² + c² = 5


Now we know that

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


Since ab + bc + ca = 10

=> -ab - bc - ca = - 10

So put the values.

=> a³ + b³ + c³ - 3abc = (5)(5 - 10)

=> a³ + b³ + c³ - 3abc = 5(-5)

=> a³ + b³ + c³ - 3abc = -25

=> a³ + b³ + c³ = - 25 + 3abc.


:) Proved.

Answer 2

$\boxed{\bold{\large{Answer:}}}$

$\text{a\:+b\:+c\:=\:5}$

$\text{ab\:+\:bc\:+\:ca\:=10}$

$\text{We\:have\:to\:find\:a^2\:+b^2\:+c^2}$

$(a + b + c {)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca \\ \\ \implies(a + b + c {)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) \\ \\ \implies(a + b + c {)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(10) \\ \\ \implies(5 {)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 20 \\ \\ \implies \: {a}^{2} + {b}^{2} + {c}^{2} = 25 - 20 = 5$

$\text{We\:know\:that,}$

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ \\ \implies5(5 - 10) \\ \\ \implies5( - 5) = - 25$

$\text{Hence\:Proved.}$

$\boxed{\bold{\large{Hope\:it\:helps\:you!}}}$