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$$2x-y = y-2z = 4$$

What is the result of $4x^2-2y^2+4z^2 $?

Factoring it and rewriting the term

$$2(2x^2-y^2+2z^2)$$

Once in a while, I've tried to factor it further. However, it is impossible to proceed anymore. Could you tell what I'm actually missing?

Regards!

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  • $\begingroup$ $8(x-z){}{}{}{}$ $\endgroup$ – Nosrati Aug 8 '18 at 12:16
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$$2x - y = 4 \Rightarrow 2x = 4 + y$$ $$y - 2z = 4 \Rightarrow 2z = y - 4$$ $$4x^2 = 16 + 8y + y^2$$ $$4z^2 = 16 - 8y + y^2$$

can you take it from here?

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