Question

Find $\sf{x^2+y^2+z^2}$ if $\sf{x^2-yz=y^2-zx=z^2-xy=i}$.
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Answer 1

Required Answer:-

Given Information:

• x² - yz = y² - xz = z² - xy = i

To find:

• The value of x² + y² + z²

Answer:

• The value of x² + y² + z² is 2i

Solution:

Given,

➡ x² - yz = y² - xz

➡ x² - y² = yz - xz

➡ x² - y² = -xz + yz

➡ (x + y)(x - y) = -z(x - y)

Now, cancelling out (x - y) from both sides, we get,

➡ x + y = -z

➡ x + y + z = 0 .....(i)

Now, note this,

1. x² - yz = i or yz = x² - i
2. y² - xz = i or xz = y² - i
3. z² - xy = i or xy = z² - i

Now, squaring both sides of equation (i), we get,

➡ x² + y² + z² + 2(xy + yz + xz) = 0

Substituting the values of xy, yz and xz, we get,

➡ x² + y² + z² + 2(z² - i + x² - i + y² - i) = 0

➡ x² + y² + z² + 2z² + 2x² + 2y² - 2i - 2i - 2i = 0

➡ 3(x² + y² + z²) = 6i

Dividing both sides by 3, we get,

➡ x² + y² + z² = 2i

Hence, the value of x² + y² + z² is 2i

Note: i = √-1 (imaginary number)

Answer 2

Given:

x² - yz = y? - xz = z² - xy = i

To find:

• The value of x? + y² + z?

Answer:

The value of x² + y? + z? is 2i

Solution:

Given,

➦x - yz = y? - xz

➦ x² - y? = yz - XZ

➦x² - y? = -xz + yz

➦(x + y)(x - y) = -z(x - y)

Now, cancelling out (x - y) from both sides, we get

x + y = -z

X+ y + z = 0

Now, note this,

1) x² - yz = i or yz = x² - i

2) y - xz = i or xz = y2 - i

3) z? - xy = i or xy = z? i

Now, squaring both sides of equation (i), we get,

x2 + y2 + z2 + 2xy + yz + xz) = 0

Substituting the values of xy, yz and xz, we get,

➦ x² + y? + z? + 2(z² - i + x² - i + y² - i) = 0

➦x² + y2 + z2 + 2z? + 2x? + 2y? - 2i - 2i - 2i

= 0

➦ 3(x? + y? + z?) = 6i

Dividing both sides by 3, we get,

x² + y² + z = 2i

Hence, the value of x? + y? + z? is 2i