IIT JEE , AIEEE Forum - Mathematics , Algebra - if x+y+z=1, x^2+y^2+z^2=2, x^3+y^3+z^3=3, then find x^4+y^4+z^4

konichiwa2x

ok here goes...

----(1)

-----(2)

----(3) (from (1) and (2))

----(4)

(from (3) and (4))

the.sniper

HEY KONICHIWA FROM WHERE DO U USE THESE SPECIAL SYMBOLS ,THEY ARE NOT OF GOIIT FORMULA EDITOR.PLEASE TELL ME I ALSO WANT THEM

hsbhatt

We have (x+y+z)(x³+y³+z³) =

x⁴+

xy(x²+y²)

xy(x²+y²) =

xy(2-z²) (Since

x² = 2.)

=2

xy-xyz

x

Hence

x⁴ = (x+y+z)(x³+y³+z³) - 2

xy+xyz

x

(

x)² =

x² + 2

xy giving

xy = -1/2

Now we have to find xyz

From the above x, y and z are solutions to the cubic

t³- t²- t/2 + c=0 where c = -xyz

Hence x³- x²- x/2 + c = 0 .........1

y³- y²- y/2 + c = 0 ...........2

z³- z²- z/2 + c = 0 ...........3

Adding the three eqns

we get

x³ -

x² -

x/2 + 3c = 0

giving c = -1/6 or xyz = 1/6

x⁴ = (x+y+z)(x³+y³+z³) - 2

xy+xyz

x = 1*3 - 2*(-1/2)+1/6 = 25/6

Corrected as pointed out by Konichiwa

konichiwa2x

there is a mistake in the last line, please correct it. The correct answer is 25/6.

for the.sniper: The software/language is called latex. google it.

nadeemoidu