Question

when a+b+c=9 and a2+b2+c2= -35 then find a3+b3+c3-3abc​

Answer 1

$\large\bf\underline \blue {To \: \mathscr{f}ind:-}$

• we need to find the value of a³ + b³ + c³ - 3abc

$\huge\bf\underline \purple{ \mathcal{S}olution:-}$

$\bf\underline{\red{Given:-}}$

• a² + b² + c² = -35
• a + b + c = 9

we know that:-

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

So,

• Value of a³ + b³ + c³ - 3abc :-

• a² + b² + c² = -35 .....1)
• a + b + c = 9 ....2)

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

• we know that,

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

↛ (a + b+ c)² -( a² + b² + c² ) = 2[ab + bc + ca]

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

↛ 9² -(-35)/2 = (ab + bc + ca)

↛ 81 + 35/2 = (ab + bc + ca)

↛ 116/2 = (ab + bc + ca)

↛ 58 = (ab + bc + ca) ......4)

• Putting values of (a + b + c) , (a² + b² + c²) , (ab +bc + ca) in 3)

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

↛ a³ + b³ + c³ - 3abc = 9 × [-35 - (58)]

↛ a³ + b³ + c³ - 3abc = 9 × [ - 93 ]

↛ a³ + b³ + c³ - 3abc = -837

Hence,

• Value of a³ + b³ + c³ - 3abc is - 837

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