x=y=333 and z=334 find the values x cube+y cube+ z cube-3xyz
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Answered by
4
x^3+y^3+z^3-3xyz =( x+y+z)(x^2+y^2+z^2-xy-yz-zx)
= (333+333+334)(333*333+333*333+334*334-333*333*-333*334-334*333)
=(1000)(
[333*333-333*333+333*333+334*334-2(333*334)]
=(1000)(0+110889+111556-2(111222))
=(1000)(222445-222444)
=1000*1
=1000
= (333+333+334)(333*333+333*333+334*334-333*333*-333*334-334*333)
=(1000)(
[333*333-333*333+333*333+334*334-2(333*334)]
=(1000)(0+110889+111556-2(111222))
=(1000)(222445-222444)
=1000*1
=1000
Answered by
7
Answer:
1000
Step-by-step explanation:
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