Question

if x=y=222 and z=223 then the value of x cube +y cube+z cube-3xyz​

Answer 1

Answer:

x³+y³+z³-3xyz

(222)³+(222)³+(223)³-3(222)(222)(223)

=10941048+10941048+11089567-32970996

=667.

Answer 2

Answer:

The value of the given expression is 667

Therefore $x^3+y^3+z^3-3xyz=667$

Step-by-step explanation:

Given that the values are x=222 ,y=222 and z=223

To find the value of $x^3+y^3+z^3-3xyz$ :

$x^3+y^3+z^3-3xyz$

• Substitute the values of x,y and z in the above expression we get

$=(222)^3+(222)^3+(223)^3-3(222)(222)(223)$

$=(222\times 222\times 222)+(222\times 222\times 222)+(223\times 223\times 223)-3(222)(222)(223)$

$=10941048+10941048+11089567-32970996$

$=32971663-32970996$

$=667$

Therefore the value of the given expression is 667

Therefore $x^3+y^3+z^3-3xyz=667$