**Santanu Sahoo **asked in Math

Question :

Prove that :

$Q.{\left(-{a}^{2}+{b}^{2}+{c}^{2}\right)}^{2}=\left(2bc-{a}^{2}+{b}^{2}+{c}^{2}\right)\left(2bc+{a}^{2}-{b}^{2}-{c}^{2}\right)$

Advika Anand answered this

Vinayak Gupta answered this

Santanu Sahoo answered this

I am the asker of this question!

**Actually, the above question has an error. The actual question is:**

Sorry for that error.

Yours Faithfully,

Santanu Sahoo (The asker of this question)

Santanu Sahoo answered this

Good Evening!

To the Experts,**I am the asker of this question. **

Actually, I tried to prove my actual question. Firstly, I took the L.C.M to be 4c^{2} and then put the law of exponents and then finally applied, a^{2} - b^{2} = (a+b)(a-b).

Using this I got my answer.

So, I would recommend not to answer this now un-necessarily.

Thanking You,

Good Night!

Sanjari Kalantri answered this

$Toprove:{b}^{2}-{\left(\frac{-{a}^{2}+{b}^{2}+{c}^{2}}{2c}\right)}^{2}=\frac{\left(2bc-{a}^{2}+{b}^{2}+{c}^{2}\right)\left(2bc+{a}^{2}-{b}^{2}-{c}^{2}\right)}{4{c}^{2}}\phantom{\rule{0ex}{0ex}}Proof:LHS={b}^{2}-{\left(\frac{-{a}^{2}+{b}^{2}+{c}^{2}}{2c}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(b+\frac{-{a}^{2}+{b}^{2}+{c}^{2}}{2c}\right)\left(b-\frac{-{a}^{2}+{b}^{2}+{c}^{2}}{2c}\right)\left[\because {x}^{2}-{y}^{2}=\left(x+y\right)\left(x-y\right)\right]\phantom{\rule{0ex}{0ex}}=\left(\frac{2bc+\left(-{a}^{2}+{b}^{2}+{c}^{2}\right)}{2c}\right)\left(\frac{2bc-\left(-{a}^{2}+{b}^{2}+{c}^{2}\right)}{2c}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(2bc-{a}^{2}+{b}^{2}+{c}^{2}\right)\left(2bc+{a}^{2}-{b}^{2}-{c}^{2}\right)}{4{c}^{2}}\phantom{\rule{0ex}{0ex}}=RHS\phantom{\rule{0ex}{0ex}}Hence,proved.$

Hope this information will clear your doubts about topic.

If you have any more doubts, please ask here on the forum and our experts will try and help you out as soon as possible.

Regards