How do you factor #a^6 + 1#?

Answer 1

We can modify the expression to use the Sum of Cubes formula to factorise it.

#a^6 + 1 = (a^2)^3 + 1^3#
The formula says :#color(blue)(x^3 + y^3 = (x + y)(x^2-xy+y^2)#
Here, #x# is #a^2# and #y# is #1#
#(a^2)^3 + 1^3 = (a^2+1){(a^2)^2 - (a^2*1)+1^2}#
#color(green)(= (a^2+1)(a^4 - a^2+1)#
As none of the factors can be factorised further, this becomes the Factorised form of #a^6 + 1#
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Answer 2

To factor (a^6 + 1), you can use the difference of squares formula, which states that (a^2 - b^2) can be factored as ((a - b)(a + b)). In this case, (a^6 + 1) can be rewritten as ((a^2)^3 + 1^3). Applying the difference of cubes formula, we have ((a^2 + 1)((a^2)^2 - a^2 + 1)). Further simplification yields ((a^2 + 1)(a^4 - a^2 + 1)). Therefore, (a^6 + 1) factors into ((a^2 + 1)(a^4 - a^2 + 1)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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