Question

P=(a+2b-c)(a^2+4b^2+c^2-2ab+2bc+ac), a, b, c is integer number satisfying 'a+2b-c=1' then, what is the rest of P/6?

Answer 1

Answer:

P = (a+2b-c) (a²+4b²+c²-2ab+2bc+ac)

Here ,

(a + 2b - c) = 1

Therefore,

P = (a²+4b²+c²-2ab+2bc+ac )

And ,

P/6 = (a²+4b²+c²-2ab+2ab+ac) / 6.

Step-by-step explanation:

Answer 2

Aɴꜱᴡᴇʀ

$\frac{p}{6} = \frac{ {a}^{2} + {4b}^{2} + {c}^{2} - 2ab + 2bc + ac }{6}$

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General knowledge

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Sᴛᴇᴘꜱ

we know the value of p

so when a+2b-c becomes one and then the only value remaining is

$p= {a}^{2} + {4b}^{2} + {c}^{2} - 2ab + 2bc + ac$

so when this is divided by 6 we get

$\frac{p}{6} = \frac{ {a}^{2} + {4b}^{2} + {c}^{2} - 2ab + 2bc + ac }{6}$

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