Question

If a+b+c=9 and a2+b2+c2=35 find the value of a3+b3+c3-3abc

Answer 1

$\textbf{Answer :}$

Given,

a + b + c = 9 ...(i)

a² + b² + c² = 35 ...(ii)

➽ (a + b + c)² - 2 (ab + bc + ca) = 35

➽ 9² - 2 (ab + bc + ca) = 35

➽ 2 (ab + bc + ca) = 81 - 35

➽ 2 (ab + bc + ca) = 46

➽ ab + bc + ca = 23 ...(iii)

Now,

a³ + b³ + c³ - 3abc

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

= (a + b + c) {(a² + b² + c²) - (ab + bc + ca)}

= 9 (35 - 23), using (i), (ii) and (iii)

= 9 × 12

= 108

#$\textbf{MarkAsBrainliest}$

Answer 2

a + b + c = 9

(a + b + c)² = 9²

a² + b² + c² + 2(ab + bc + ca) =81

35 + 2(ab + bc + ca) 81

2(ab + bc + ca) = 81 - 35 =46

ab + bc + ca = 46/2 = 23

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We know,

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

Putting the given value,

(9)[35-1(23)]

9(35-23)

9(12)

108

I hope this will help you

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