Question

If a+b+c=9 and ab+bc+ca=26, find the value of a³ +b³ +c³ -3abc.

Answer 1

Given : a + b + c = 9, ab + bc + ca = 26

On Squaring, a + b + c = 9 both sides, we get

(a + b + c)² = 9²

a² + b² + c² + 2 (ab + bc + ca) = 81

[By using an identity, (a + b + c)² = a² + b² + c² + 2ab + 2 bc + 2 ca)]

a² + b² + c² + 2 x 26 = 81

a² + b² + c² + 52 = 81

a² + b² + c² = 81 - 52

a² + b² + c² = 29………..(1)

Now, by using an identity , a³ + b³ + c³ - 3abc = (a + b + c) [(a² + b² + c² – ab -  bc - ca)]

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

= 9[29 – 26]

[Given : a + b + c = 9, ab + bc + ca = 26 and from eq1]

= 9 x 3

= 27

a³ + b³ + c³ - 3abc  = 27

Hence the value of a³ + b³ + c³ - 3abc is 27.

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Answer 2

Answer:

Step-by-step explanation:

Given : a + b + c = 9, ab + bc + ca = 26

On Squaring, a + b + c = 9 both sides, we get

(a + b + c)² = 9²

a² + b² + c² + 2 (ab + bc + ca) = 81

[By using an identity, (a + b + c)² = a² + b² + c² + 2ab + 2 bc + 2 ca)]

a² + b² + c² + 2 x 26 = 81

a² + b² + c² + 52 = 81

a² + b² + c² = 81 - 52

a² + b² + c² = 29………..(1)

Now, by using an identity , a³ + b³ + c³ - 3abc = (a + b + c) [(a² + b² + c² – ab -  bc - ca)]

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

= 9[29 – 26]

[Given : a + b + c = 9, ab + bc + ca = 26 and from eq1]

= 9 x 3

= 27

a³ + b³ + c³ - 3abc  = 27