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The HCF of $({{x}^{2}}-9),\,({{x}^{3}}-27)\:\text{and}\:({{x}^{2}}-8x+15)$ is
\[\text{A}.\,\,\,x-3\]
$\text{B}.\,\,\,x+3$
$\text{C}.\,\,\,(x-3)({{x}^{2}}-3x+9)$
$\text{D}.\,\,\,(x-3)(x+3)(x+5)({{x}^{2}}+3x+9)$

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Hint: find the factor or rather factorize them. The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula.
Formula and
\[\begin{align}
  & {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \\
 & {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}-ab+{{b}^{2}}) \\
 & {{a}^{3}}+{{b}^{3}}=(a-b)({{a}^{2}}-ab+{{b}^{2}}) \\
\end{align}\]
Also we can add ${{a}^{3}}-{{b}^{3}}={{(a-b)}^{3}}+3{{a}^{2}}b-3a{{b}^{2}}$

Complete step by step solution: In mathematics, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind. For example, 3 × 5 is a factorization of the integer 15, and is a factorization of the polynomial x² – 4.
$({{x}^{2}}-9);\,({{x}^{3}}-27)\,.\,\text{and}\,({{x}^{2}}-8x+15).({{x}^{2}}-9)$
So, this is the given part to which we have to find the Hcf
So, using the formula $[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)]$
We have
\[\begin{align}
  & {{x}^{2}}-9 \\
 & \Rightarrow {{x}^{2}}-{{3}^{2}}=(x-3)(x+3) \\
\end{align}\]
Also, using the formula ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}-ab+{{b}^{2}})$
We have

\[\begin{align}
  & {{x}^{3}}-27 \\
 & \Rightarrow {{x}^{3}}-33=(x-3)({{x}^{2}}-2x+3) \\
\end{align}\]
Now, if we factorize ${{x}^{2}}-8x+15$
We get
\[\begin{align}
  & {{x}^{2}}-8x+15={{x}^{2}}-5x-3x+15 \\
 & =x(x-5)-3(x-5) \\
 & =(x-3)(x-5)
\end{align}\]
Therefore, we can see $(x-3)$is the only common factor in $({{x}^{2}}-9),\,({{x}^{2}}-27)({{x}^{2}}-8x+15)$
Hence the HCF for this question is $(x-3)$

Therefore option (A) is the correct answer.

Note: In this type of question, factorization is must know, find factors then see for the common factors and as HCF suggest the common factor is the HCF of the question. The highest common factor is found by multiplying all the factors which appear in both lists: So the HCF of 60 and 72 is 2 × 2 × 3 which is 12. The lowest common multiple is found by multiplying all the factors which appear in either list: So the LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 which is 360.
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