Question #8db74

Question

Question #8db74

2 Answers

Impact of this question

1071 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-23T07:55:22

$? = x = \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}$ $?=x=1/a^2+ 1/b^2+ 1/c^2$

Explanation:

If $b \cdot c + c \cdot a + a \cdot b = 0$ $b*c + c*a + a*b = 0$ ,

Prove that $\frac{1}{a^{2}} - b \cdot c + \frac{1}{b^{2}} - c \cdot a + \frac{1}{c^{2}} - a \cdot b = x$ $1/a^2 -b*c + 1/b^2 - c*a + 1/c^2 -a*b = x$ .

Solve for x, I presume this is what the problem is you are trying to solve. This is not really something you actually call a proof, since you are solving for an unknown...

We can rewrite $\frac{1}{a^{2}} - b \cdot c + \frac{1}{b^{2}} - c \cdot a + \frac{1}{c^{2}} - a \cdot b = x$ $1/a^2 -b*c + 1/b^2 - c*a + 1/c^2 -a*b = x$ this to:
$\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = x + b \cdot c + c \cdot a + a \cdot b$ $1/a^2+ 1/b^2+ 1/c^2 = x + b*c + c*a + a*b$
$\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = x + 0$ $1/a^2+ 1/b^2+ 1/c^2 = x + 0$
So, $x$ $x$ or your question mark, $? = x = \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}$ $?=x=1/a^2+ 1/b^2+ 1/c^2$ .

Answer 2 · 2018-01-23T07:57:53

Image reference....

Explanation:

![my http://notebook....](https://useruploads.socratic.org/zyyW3GBSRmOfEtPXpiGQ_New%20Doc%202018-01-23_1.jpg)

Hope it helps....
Thank you

Science

Math

Humanities

... and beyond

Question #8db74

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question