Question 42

If$$x^{2}+ y^{2}+ z^{2} $$= xy + yx + zx, then the value of $$\frac{3x^{4}+7y^{4}+5z^{4}}{5x ^{2}y^{2}+7y ^{2}z^{2}+3z^{2}x^{2}}$$

Solution

We know that if $$x^{2}+ y^{2}+ z^{2} $$= xy + yx + zx, Then x=y=z

Therefore, substituting x=y=z in given expression $$\frac{3x^{4}+7y^{4}+5z^{4}}{5x ^{2}y^{2}+7y ^{2}z^{2}+3z^{2}x^{2}}$$, we get

= $$\frac{15x^4}{15x^4}$$

=1

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: