Question

if x+y+z=0 then x cube +y cube + z cube =​

Answer 1

Required Answer:-

Given:

• x + y + z = 0

To find:

• x³ + y³ + z³ = ?

Answer:

• x³ + y³ + z³ = 3xyz.

Solution:

We have,

➡ x + y + z = 0

➡ x + y = -z

So,

x³ + y³ + z³

= (x³ + y³) + z³

Now, we know that,

➡ (x³ + y³) = (x + y)³ - 3xy(x + y)

So,

(x³ + y³) + z³

= (x + y)³ - 3xy(x + y) + z³

Substitute x + y = -z here, we get,

= -z³ - 3xy × (-z) + z³

= -3xy × (-z)

= 3xyz

Hence,

➡ x³ + y³ + z³ = 3xyz.

Answer 2

Question:-

• If x+y+z=0
• Find: ${x}^{3} + {y}^{3} + {z}^{3} = ?$

Answer:-

Given:

• x+y+z=0

We know,

$\sf(x + y + z) ( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz) = {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

But (x+y+z)=0

$\therefore \sf(0) ( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz) = {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

$\implies \sf \: 0 = {x}^{3} + {y}^{3} + {z}^{3} - 3xyz \\ \implies\sf 3xyz = {x}^{3} + {y}^{3} + {y}^{3}$

Correct Answer:-

• 3xyz

Learn more about Algebric identity

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$