Question

Please solve the question immediately​

Answer 1

Answer:

a^4+b^4+c^4=46

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Answer 2

SOLUTION :

According to question:;

a0x3+a1x²+a2x+A3=0 ( 1 )

Then, s1 =4, S2 =10 & S3= 22

Hence,

=a0 s1 + a1 = 0

=4a0 +a1 = 0

then, a1= -4a0

=a0 s2 + a1 s1 + 2a2 = 0

or, a0 10 - 4a0 × 4+ 2a2 = 0

or, 10a0 - 16a0 + 2a2 = 0

or, -6a0 +2a2 = 0

or, 2 a2 =6a0

or, a2 = 6a0/2

Then a2 = 3a0

= a0 s 3+ a2 s1 + a1 s2 +3 a3 = 0

or, 22 a0 +3a0 × 4 -4a0 × 10 + 3 a3 = 0

or, 22 a0 +12 a0 -40 a0 + 3 a3= 0

or, -6 a0 + 3 a3 = 0

or, 3 a3 = 6 a0

Then a3 = 2a0

PUTTING THE VALUE IN EQUATION NUMBER. (1)

or, x4 - 4 x3+ 3 x2 +2 x2 = 0

CHANGE THE x INTO s

THEN :

or , s4- 4 s 3 + 3 s2 + 2s = 0

or, s4- 4 × 22+ 3 × 10+ 2 × 4 = 0

or , s4- 88+ 30 + 8= 0

or,. s4 - 50 = 0

or, s4 = 50

HENCE ::

a4 + b4 + c4 = 50