9b²c²+18c²a²+8a²b²-a⁴-16b⁴-81c⁴
Factorise it 

To factorize the expression \(9b^2c^2 + 18c^2a^2 + 8a^2b^2 - a^4 - 16b^4 - 81c^4\), let's proceed by rearranging and using grouping techniques. Here’s a step-by-step solution:

### Step 1: Arrange terms in a recognizable form
Rewrite the expression by grouping terms that might be factors of each other:
\[
9b^2c^2 + 18a^2c^2 + 8a^2b^2 - a^4 - 16b^4 - 81c^4
\]

### Step 2: Look for patterns
Notice that the terms \(9b^2c^2\), \(18a^2c^2\), and \(8a^2b^2\) are cross-terms of the variables \(a\), \(b\), and \(c\). The other terms (\(-a^4\), \(-16b^4\), \(-81c^4\)) look like squares.

Let’s rewrite each term as:
\[
9b^2c^2 = (3bc)^2, \quad 18a^2c^2 = 2 \cdot 3ac \cdot 3bc, \quad 8a^2b^2 = (2ab)^2
\]
and
\[
a^4 = (a^2)^2, \quad 16b^4 = (4b^2)^2, \quad 81c^4 = (9c^2)^2.
\]

### Step 3: Recognize it as a quadratic in terms of \( (a^2, b^2, c^2) \)
The expression is equivalent to:
\[
(3bc + 3ac + 2ab)^2 - (a^2 + 4b^2 + 9c^2)^2
\]

Thus,

9b²c² + 18c²a² + 8a²b² - a⁴ - 16b⁴ - 81c⁴. Factorise it.

Explore how to factorize the polynomial 9b²c² + 18c²a² + 8a²b² - a⁴ - 16b⁴ - 81c⁴ using algebraic techniques. This solution demonstrates methods including grouping, recognizing squares, and the difference of squares theorem. Ideal for high school students in Grades 10-12.

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