Practicing Success

Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Algebra

Question:

If x + y + z = 3, $x^2 + y^2 + z^2 = 45$ and $x^3 + y^3 + z^3 = 69,$ then what is the value of xyz ?

Options:

-40

40

-31

30

Correct Answer:

-40

Explanation:

x + y + z = 3,

$x^2 + y^2 + z^2 = 45$

$x^3 + y^3 + z^3 = 69,$

then what is the value of xyz= ?

We know that, (x + y + z)2 =  x2 + y2 + z+ 2(xy + yz + zx)

 x3 + y3 + z3 – 3xyz = (x + y + z){( x2 + y2 + z) – (xy + yz +zx)}

Now,

= (x + y + z)2 =  x2 + y2 + z + 2(xy + yz + zx)

= (3)2 = 45 + 2(xy + yz +zx)

= 9 = 45 + 2(xy + yz +zx)

= -36 = 2(xy + yz +zx)

= (xy + yz +zx) = -18

Now,

x3 + y3 + z– 3xyz = (x + y + z){( x2 + y2 + z) – (xy + yz +zx)}

= 69 – 3xyz = 3 {(45) – (-18)}

= 69 – 3xyz = 3 × {(45 + 18)}

= 69 – 3xyz = 3 × 63

= 69 – 3xyz = 189

= 3xyz = (69 – 189)

= 3xyz = -120

= xyz = -40