Question 20

$$A = \frac{(x^8 - 1)}{(x^4 + 1)}$$ and $$B = \frac{(y^4 - 1)}{(y^2 + 1)}$$. If $$x = 2$$ and $$y = 9$$, then what is the value of $$A^2 + 2AB + A(A+B)^2$$?

Solution

$$A = \frac{(x^8 - 1)}{(x^4 + 1)}$$ and $$B = \frac{(y^4 - 1)}{(y^2 + 1)}$$
$$A = (x^4 - 1)$$
$$B=(y^2 - 1)$$
For x=2 ,A=15
For y=9,B=80
$$A^2 + 2AB + A(A+B)^2$$=225+2400+(15)(9025)
=2625+135375
=138000

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SSC CGL Tier-2 18-February-2018 Maths

$$A = \frac{(x^8 - 1)}{(x^4 + 1)}$$ and $$B = \frac{(y^4 - 1)}{(y^2 + 1)}$$. If $$x = 2$$ and $$y = 9$$, then what is the value of $$A^2 + 2AB + A(A+B)^2$$?

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