Question : A complete factorisation of $(x^4+64)$ is:
Option 1: $(x^2+8)^2$
Option 2: $(x^2+8)$$(x^2-8)$
Option 3: $(x^2-4x+8)(x^2-4x-8)$
Option 4: $(x^2+4x+8)$$(x^2-4x+8)$
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Correct Answer: $(x^2+4x+8)$$(x^2-4x+8)$
Solution : Given: $(x^4+64)$ We know that the algebraic identities are $(x^2+y^2)=(x+y)^2-2xy$ and $(x^2-y^2)=(x+y)(x-y)$. So, $(x^4+64)=(x^2)^2+8^2$ $=(x^2+8)^2-2x^2\times8$ $=(x^2+8)^2-(4x)^2$ $=(x^2+4x+8)(x^2-4x+8)$ Hence, the correct answer is $(x^2+4x+8)(x^2-4x+8)$.
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Question : Simplify the following expression. $(4 x + 1)^2 - (4 x + 3)(4 x - 1)$
Option 1: $(4 x + 1)$
Option 2: $4$
Option 3: $(4x- 3)$
Option 4: $4x$
Question : If $(x–4)(x^2+4x+16)=x^3–p$, then $p$ is equal to:
Option 1: 27
Option 2: 8
Option 3: 64
Option 4: 0
Question : What is the LCM of $\left(8 x^3+80 x^2+200 x\right)$ and $\left(4 x^4+16 x^3-20 x^2\right)$?
Option 1: $8 x^2(x+5)^2(x-1)$
Option 2: $8 x^2(x-1)^2(x+5)$
Option 3: $4 x^2(x-1)^2(x+5)$
Option 4: $4 x^2(x+5)^2(x-1)$
Question : If $x^2-8 x-1=0$, what is the value of $x^2+\frac{1}{x^2}?$
Option 1: 68
Option 2: 62
Option 4: 66
Question : If $x=a+\frac{1}{a}$ and $y=a-\frac{1}{a}$, then the value of $x^4+y^4-2x^2y^2$ is:
Option 1: 4
Option 3: 16
Option 4: 64
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