A-c-b-c- - -a-b-2c-a-b-2c-a-b-a-b-

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a-c+(b-c) = _...

Question

a-{c+(b-c)} = _____.

a-b-2c

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a+b-2c

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a-b

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a+b

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Similar questions

a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a}

If a, b, c are in G.P., prove that:

(i) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

(iii) $\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}} = \frac{a + b + c}{a - b + c}$

(iv) $\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}$

(v) $(a + 2 b + 2 c) (a - 2 b + 2 c) = a^{2} + 4 c^{2}$ .

a, b, c

\frac{1}{2} {log}_{e} a + \frac{1}{2} {log}_{e} c + (\frac{1}{2 c a + 1}) + \frac{1}{3} {(\frac{1}{2 c a + 1})}^{3} + \frac{1}{5} {(\frac{1}{2 c a + 1})}^{5} + . . . \infty = {log}_{e} b

b

(2 a + b + c) (4 a^{2} + b^{2} + c^{2} - 2 a b - b c - 2 c a)

Prove that

2a^2+2b^2+2c^2-2ab-2bc-2ca={(a-b)^2+(b-c)^2+(c-a)^2}

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