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What is the factored form of [tex][tex]$r^{27}-s^{30}$[/tex][/tex]?

A. [tex]\left(r^9 - s^{10}\right)\left(r^9 - s^{10}\right)\left(r^9 - s^{10}\right)[/tex]

B. [tex]\left(r^9 - s^{10}\right)\left(r^{18} + r^9 s^{10} + s^{20}\right)[/tex]

C. [tex]\left(r^9 - s^{10}\right)\left(r^9 + r^9 s^{10} + s^{10}\right)[/tex]

D. [tex]\left(r^9 - s^{10}\right)\left(r^{18} + r^{18} s^{20} + s^{20}\right)[/tex]


Sagot :

To factor the expression [tex]\( r^{27} - s^{30} \)[/tex], one can use the properties of polynomials and factorization.

Let's start by observing that both terms [tex]\( r^{27} \)[/tex] and [tex]\( s^{30} \)[/tex] have powers that are multiples of 3. We can then factor this as a difference of cubes. Specifically, we note that:

[tex]\[ r^{27} = (r^9)^3 \][/tex]
[tex]\[ s^{30} = (s^{10})^3 \][/tex]

By the difference of cubes formula, [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex], we identify [tex]\( a = r^9 \)[/tex] and [tex]\( b = s^{10} \)[/tex].

Applying the formula:

[tex]\[ r^{27} - s^{30} = (r^9)^3 - (s^{10})^3 \][/tex]
[tex]\[ = (r^9 - s^{10})((r^9)^2 + r^9(s^{10}) + (s^{10})^2) \][/tex]

Simplifying inside the second bracket:

[tex]\[ (r^9)^2 = r^{18} \][/tex]
[tex]\[ r^9 \cdot s^{10} = r^9 s^{10} \][/tex]
[tex]\[ (s^{10})^2 = s^{20} \][/tex]

Hence, we get:

[tex]\[ r^{27} - s^{30} = (r^9 - s^{10})(r^{18} + r^9 s^{10} + s^{20}) \][/tex]

Thus, the factored form of [tex]\( r^{27} - s^{30} \)[/tex] is:

[tex]\[ (r^9 - s^{10})(r^{18} + r^9 s^{10} + s^{20}) \][/tex]

Therefore, the correct answer is:

[tex]\[ \boxed{\left(r^9 - s^{10}\right)\left(r^{18} + r^9 s^{10} + s^{20}\right)} \][/tex]