Question

Please refer to attachment below — having a hard time answering this one.

Answer 1

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)

=> (5)[a² + b² + c² -(ab + bc + ca)]

Again , a² + b² + c² = (a + b + c)² - 2(ab +

bc + ca)

Substitute the values,

=> a² + b² + c² = (5)² - 2(10) = 25 - 10

=> a² + b² + c² = 5

=> a³ + b³ + c³ - 3abc = 5(5 - 10)

=> a³ + b³ + c³ - 3abc = - 25

Hence, proved.

Answer 2

$\huge {\bigstar{\mathfrak\red {Answer:}}}$

Step-by-step explanation:

$We \: are \: given, \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: a + b + c = 5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \:ab + bc + ca = 10 \\ \\ a + b + c = 5 \\ ⇒ {(a + b + c)}^{2} = {5}^{2} \\ ⇒{a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 25 \\ ⇒ {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 25 \\ ⇒ {a}^{2} + {b}^{2} + {c}^{2} + 2 \times 10 = 25 \\ ⇒ {a}^{2} + {b}^{2} + {c}^{2} + 20 = 25 \\ ⇒ {a}^{2} + {b}^{2} + {c}^{2} = 25 - 20 \\ ⇒ {a}^{2} + {b}^{2} + {c}^{2} = 5 \\ \\A s \: we \: know \: that, \\ \\ {a}^{3} + {b}^{3} + {c}^{3} - 3 abc = ( a+ b+ c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ ⇒ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)({a}^{2} + {b}^{2} + {c}^{2} - (ab + bc + ca)) \\ ⇒ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 5 \times (5 - 10) \\ ⇒ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 5 \times ( - 5) \\∴ {a}^{3} + {b}^{3} + {c}^{3} - abc = - 25 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [Hence \: Proved ]$

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