Simplifying Complex Algebraic Expressions

The problem asks us to simplify a given algebraic expression which is in the form of a fraction. The expression involves cubes of binomial and trinomial terms.

The given expression is:

\[{{{({{a}^{2}}-4{{b}^{2}})}^{3}} + 64 {{(b^2 - 4c^2)}^3} + {{(16c^2 -a^2})^3}} \over {{(a-2b)^3} +{(2b-4c)^3} +{(4c-a)^3}}\]

We can simplify this expression by recognizing and applying the algebraic identity: If \(x + y + z = 0\), then \(x^3 + y^3 + z^3 = 3xyz\).

Step 1: Analyze the Denominator

Let's examine the terms in the denominator:

• Let \(x = a - 2b\)
• Let \(y = 2b - 4c\)
• Let \(z = 4c - a\)

Now, let's check the sum of these terms:

\[x + y + z = (a - 2b) + (2b - 4c) + (4c - a)\]

Group like terms:

\[x + y + z = (a - a) + (-2b + 2b) + (-4c + 4c) = 0 + 0 + 0 = 0\]

Since \(x + y + z = 0\), we can apply the identity \(x^3 + y^3 + z^3 = 3xyz\) to the denominator.

Denominator \( = (a-2b)^3 + (2b-4c)^3 + (4c-a)^3 = 3(a-2b)(2b-4c)(4c-a)\).

We can factor out a 2 from the term \((2b - 4c)\):

\[2b - 4c = 2(b - 2c)\]

So, the denominator becomes:

\[3(a-2b) \cdot 2(b-2c) \cdot (4c-a) = 6(a-2b)(b-2c)(4c-a)\]

Step 2: Analyze the Numerator

The numerator is \({{(a^2-4b^2)}^3} + 64 {{(b^2 - 4c^2)}^3} + {{(16c^2 -a^2})^3}}\).

Notice that \(64 = 4^3\). So, the term \(64 {{(b^2 - 4c^2)}^3}\) can be written as \((4{(b^2 - 4c^2)})^3\).

Let's define the terms being cubed in the numerator (after adjusting the constant):

• Let \(A = a^2 - 4b^2\)
• Let \(B = 4(b^2 - 4c^2)\)
• Let \(C = 16c^2 - a^2\)

Now, let's check the sum of these terms:

\[A + B + C = (a^2 - 4b^2) + 4(b^2 - 4c^2) + (16c^2 - a^2)\]

\[A + B + C = a^2 - 4b^2 + 4b^2 - 16c^2 + 16c^2 - a^2\]

Group like terms:

\[A + B + C = (a^2 - a^2) + (-4b^2 + 4b^2) + (-16c^2 + 16c^2) = 0 + 0 + 0 = 0\]

Since \(A + B + C = 0\), we can apply the identity \(A^3 + B^3 + C^3 = 3ABC\) to the numerator.

Numerator \( = A^3 + B^3 + C^3 = 3ABC\)

Substitute back the expressions for A, B, and C:

\[\text{Numerator} = 3(a^2 - 4b^2)(4(b^2 - 4c^2))(16c^2 - a^2)\]

Rearrange the constant:

\[\text{Numerator} = 3 \cdot 4 \cdot (a^2 - 4b^2)(b^2 - 4c^2)(16c^2 - a^2)\]

\[\text{Numerator} = 12(a^2 - 4b^2)(b^2 - 4c^2)(16c^2 - a^2)\]

We can further factor the terms using the difference of squares formula \((p^2 - q^2 = (p-q)(p+q))\):

• \(a^2 - 4b^2 = a^2 - (2b)^2 = (a - 2b)(a + 2b)\)
• \(b^2 - 4c^2 = b^2 - (2c)^2 = (b - 2c)(b + 2c)\)
• \(16c^2 - a^2 = (4c)^2 - a^2 = (4c - a)(4c + a)\)

So, the numerator becomes:

\[\text{Numerator} = 12(a - 2b)(a + 2b)(b - 2c)(b + 2c)(4c - a)(4c + a)\]

Step 3: Simplify the Fraction

Now, we have the simplified numerator and denominator:

\[\text{Expression} = {{\text{Numerator}} \over {\text{Denominator}}}\]

\[\text{Expression} = {{12(a - 2b)(a + 2b)(b - 2c)(b + 2c)(4c - a)(4c + a)} \over {6(a-2b)(b-2c)(4c-a)}}\]

Cancel the common terms in the numerator and the denominator:

• Cancel \((a - 2b)\) from numerator and denominator.
• Cancel \((b - 2c)\) from numerator and denominator.
• Cancel \((4c - a)\) from numerator and denominator.
• Cancel the numerical factor: \(12 \div 6 = 2\).

The remaining terms are:

\[2(a + 2b)(b + 2c)(4c + a)\]

Conclusion

The simplified expression is \(2(a + 2b)(b + 2c)(4c + a)\).

Let's check this result against the given options:

The simplified expression matches Option 2.

Revision Table: Simplifying Algebraic Expressions

Additional Information: Algebraic Identities and Simplification

Algebraic identities are equations that are true for all possible values of the variables involved. They are fundamental tools for simplifying expressions, solving equations, and proving other mathematical statements. The identity \(x^3 + y^3 + z^3 = 3xyz\) when \(x+y+z=0\) is a special case of the sum of cubes identity \(x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)\).

Simplifying algebraic expressions makes them easier to understand and work with. It often involves factoring, expanding, combining like terms, and using identities. In fractional expressions, factorization is particularly useful for cancelling common factors in the numerator and the denominator.

The difference of squares identity is another common and powerful tool for factorization. Recognizing perfect squares (\(a^2\), \(4b^2=(2b)^2\), \(16c^2=(4c)^2\)) is key to applying this identity effectively.

Problems like this one test the ability to recognize patterns and apply the correct algebraic identities strategically. Breaking down the problem into simplifying the numerator and the denominator separately is a useful approach.