Question

if a, b and c are such that (a + b + c) =2
(a² + b² + c²) =6
a³ + b³ + c³ =8
then find a⁴ + b⁴ + c⁴ =???

Answer 1

solution:-

a + b + c = 2
a⁴ + a³b + a³c = 2a³
b⁴ + ab³ + b³c = 2b³
c⁴ + bc³ + ac³ = 2c³

Adding:

X + ab(a²+ b²) + ac(a² + c²) + bc(b² + c²) (where X = a⁴ + b⁴ + c⁴) = 2 x 8 = 16
X + ab(6 - c²) + ac(6 - b²) + bc(6 - a²) =16
X + 6(ab + ac + bc) – abc(c + b + a) = 16
X + 6(ab + ac + bc) – 2abc = 16.... ( 1 )

(a + b + c)² = 4 = a² + b² + c² + 2(ab + bc + ca) = 6 + 2(ab + bc + ca)
ab + bc + ca = -1....( 2 )

(a + b + c)³ = 8 = 6(a + b + c) + 2{3abc + a²( b + c) + b²(c + a) +
c²(a + b)} = 12 + 6abc + 2(2a ²- a³ + 2b² - b³ + 2c² - c³) = 12 + 6abc + 8

→abc = -2....( 3 )


Substituting (2 ) and (3 ) in (1 ): X = 16 – 6 x (-1) + 2(-2) = 18.

Answer 2

Answer:

a+b+c = 2 , square both sides , a^2 + b^2 + c^2 +2(ab+bc+ca) = 4

6 + 2(ab+bc+ca) = 4 , ab+bc+ca = -2/2 , ab+bc+ca = (-1)

We know the identity that a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca)

Putting the values given , we get

8 - 3abc = (2)[6 - (ab +bc+ ca) ] , From we get the value of abc = (-2)

Now we can write a^4 + b^4 + c^4 as

( a^2 + b^2 + c^2)^2 - 2[ (ab)^2 + (bc)^2 + (ca)^2 ]

And (ab)^2 + (bc)^2 + (ca)^2 can be written as

(ab+bc+ca)^2 - 2(ab×bc + bc×ca + ca×ab)

= (ab +bc +ca )^2 -2[ abc (a+b+c) ]

Now putting all the values we get (ab)^2 + (bc)^2 + (ca)^2 = (-1)^2 -2 ( -2 × 2) = 9

And a^4 + b^4 + c^4 = (6)^2 - 2(9) =36 - 18

Step-by-step explanation:

Hope it helps you sir..