Question

is act! sa factors a
factors Ofesthe polynomial
0c 1997 + x
ta
$if \: x + y + z = 6 \: and \: {x}^{2} + {y }^{2} + {z}^{2} = 18 \: then \: find \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$
​

Answer 1

Step-by-step explanation:

this question's answer is 54

Answer 2

Required Answer:-

Given:

• x + y + z = 6.
• x² + y² + z² = 18

To Find:

• The value of x³ + y³ + z³ - 3xyz

Solution:

Note the identity to be used here:

→ x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz)

→ (x + y + z)² = x² + y² + z² + 2(xy + yz + xz)

Now, let's solve.

Given that,

→ x + y + z = 6

→ x² + y² + z² = 18

→ (x + y + z)² = 6² [Squaring both sides]

→ x² + y² + z² + 2(xy + yz + xz) = 36

∵ x² + y² + z² = 18

→ 18 + 2(xy + yz + xz) = 36

→ 2(xy + yz + xz) = 18

→ xy + yz + xz = 9

Therefore,

x³ + y³ + z³ - 3xyz

= (x + y + z)(x² + y² + z² - xy - yz - xz)

= (x + y + z)[x² + y² + z² - (xy + yz + xz)]

= 6 × (18 - 9)

= 6 × 9

= 54.

→ x³ + y³ + z³ - 3xyz = 54

Note: If you don't know the first identity, solve this problem in this way, (given below)

Given that,

→ x + y + z = 6

→ x² + y² + z² = 18

So,

x³ + y³ + z³ - 3xyz

= (x³ + y³) + z³ - 3xyz

Now, we know that,

→ x³ + y³ = (x + y)³ - 3xy(x + y)

Therefore,

= (x³ + y³) + z³ - 3xyz

= (x + y)³ - 3xy(x + y) + z³ - 3xyz

= (x + y)³ + z³ - 3xy(x + y) - 3xyz

= [(x + y)³ + z³] - 3xy(x + y + z)

Applying the same identity here, we get,

= (x + y + z)[(x + y)² - z(x + y) + z²] - (x + y + z) × (3xy)

= (x + y + z)[(x² + 2xy + y² - xy - yz + z²) - 3xy]

= (x + y + z)[x² + 2xy - 3xy + y² + z² - xy - yz]

= (x + y + z)[x² + y² + z² - xy - yz - xz]

= 6 × [18 - (xy - yz - xz)]

Now,

→ (x + y + z)² = 6² (Squaring both sides)

→ x² + y² + z² + 2(xy + yz + xz) = 36

∵ x² + y² + z² = 18

→ 18 + 2(xy + yz + xz) = 36

→ 2(xy + yz + xz) = 18

→ xy + yz + xz = 9

Therefore,

6 × [18 - (xy + yz + xz)]

= 6 × (18 - 9)

= 6 × 9

= 54

Answer:

• x³ + y³ + z³ - 3xyz = 54.