x²+y²+z²-xy+xz+yz by x+y-z
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x2+y2+z2≥xy+yz+zx∣⋅2⇔2x2+2y2+2z2≥2xy+2yz+2zx⇔x2+x2+y2+y2+z2+z2−2xy−2yz−2zx≥0⇔⇔(x2−2xy+y2)+(y2−2yz+z2)+(z2+2zx+x2)≥0⇔⇔(x−y)2+(y−x)2+(z−x)2≥0 (A)
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Answer:
Hello mate! I guess it question is (x+y+z)(x²+y²+z²+2xy-xz-yz)
Step-by-step explanation:
(x+y+z)(x²+y²+z²+2xy-xz-yz), clearly define that the formula is {(a+b)³= (a+b)(a²-2ab+b²)}
hence your answer is x³+y³+z³.
Hope it helps ;)
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