Recall the identities !!

Algebra Level pending

x + y + z =1

$x^{2}+y^{2}+z^{2}$ = 2

$x^{3}+y^{3}+z^{3}$ = 3

Find xyz .The answer will be of the form $\frac{a}{b}$ . Enter the answer as $\frac{a+b}{ab}$ .

(correct to three decimal places )

2 solutions

Nihar Mahajan
Jan 30, 2015

let S1 = x + y + z =1

S2 = x^2 + y^2 + z^2 = 2

S3 = x^3 + y^3 + x^3 = 3

t2 = xy + yz + xz

t3 = xyz

We have ,

S2 = S1 ^ 2 - ( 2 * t2 )

=> 2 = 1 - ( 2 * t2)

=> 1 = -2 * t2

=> t2 = -1 / 2

We also have ,

S3 = S1 ^ 3 - ( 3 * S1 * t2 ) + 3(t3)

=> 3 = 1 - [ 3 * 1 * ( -1 / 2 ) ] + 3 * xyz

=> 2 = 3/2 + 3xyz

=> 1 / 2 = 3xyz

=> xyz = 1/6 ........ => a = 1 , b = 6

(a+b) / ab = 7 / 6 = 1.167 :)

Lu Chee Ket
Jan 30, 2015

Applying SPIN identities, where S = x + y + z, P = x y z , I = (1/ x + 1/ y + 1/ z) for N2 and N3:

S = 1

S^2 - 2 P I = 2

S^3 - 3 P I S + 3 P = 3

1 - 2 P I = 2 => P I = -1/ 2

1 - 3 P I + 3 P = 3 => P = (3 - 1 + 3 P I)/ 3 = 2/ 3 - 1/ 2 = 1/ 6

(1 + 6)/ (1 x 6) = 7/ 6 = 1.167 {correct to 3 decimal places}