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Question Number 207919 by efronzo1 last updated on 30/May/24

$$\downharpoonleft \underset{―}{}$$

Commented by Frix last updated on 30/May/24

$$a^2+b^2+c^2=0\vee 15$$

Answered by mr W last updated on 30/May/24

$$a=b=c=0isasolution$$$$fora\ne 0,b\ne 0,c\ne 0:$$$$\left(ii\right)\times \left(iii\right):$$$$bc=bc(a+2)(a-2)$$$$\Rightarrow (a+2)(a-2)=1\Rightarrow a^2=5$$$$a=b^2(a-2)$$$$\Rightarrow b^2=\frac{a}{a-2}=a(a+2)=a^2+2a$$$$c^2=b^2{(a-2)}^2=a(a-2)=a^2-2a$$$$a^2+b^2+c^2=a^2+a^2+2a+a^2-2a=3a^2$$$$=15$$$$\Rightarrow a^2+b^2+c^2=0or15$$

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