Common factor of \(6x^2y,\ 9xy^2,\ -15xy\) is \(3xy\).

\[6x^2y + 9xy^2 – 15xy = 3xy(2x + 3y – 5)\]

Common factor is \(a^2\).

\[a^2 – a^2b – a^2b^2 = a^2(1 – b – b^2)\]

Common factor of \(8a^4b,\ 18a^3b^2,\ 24a^2b^3,\ 2ab^4\) is \(2ab\).

\[8a^4b + 18a^3b^2 – 24a^2b^3 + 2ab^4 = 2ab(4a^3 + 9a^2b – 12ab^2 + b^3)\]

Treat \((x+3)\) as a common factor.

\[(x+3)^2 – 3(x+3) = (x+3)\big[(x+3) – 3\big] = (x+3)(x)\]

Common factor is \(2(3x+4y)\).

\[2(3x+4y)^2 – 6(3x+4y) = 2(3x+4y)\big[(3x+4y) – 3\big]\]

Common factor of \(x^2y^2,\ x^3y^3,\ 2xy\) is \(xy\).

\[x^2y^2 – x^3y^3 + 2xy = xy(xy – x^2y^2 + 2)\]

Group terms: \((2ax + bx) + (2ay + by)\)

\[= x(2a+b) + y(2a+b)\]

\[= (2a+b)(x+y)\]

Rearrange and group: \((x^3 + x) – (5x^2 + 5)\)

\[= x(x^2+1) – 5(x^2+1)\]

\[= (x^2+1)(x-5)\]

Group terms: \((a^2 + ab) + (4a + 4b)\)

\[= a(a+b) + 4(a+b)\]

\[= (a+b)(a+4)\]

Rearrange: \((3ax + 4bx) – (6ay + 8by)\)

\[= x(3a+4b) – 2y(3a+4b)\]

\[= (3a+4b)(x-2y)\]

This fits the identity \(a^2 + b^2 + 2ab = (a+b)^2\).

Let \(a = \dfrac{4x^2}{7y}\) and \(b = \dfrac{7y^2}{4x}\).

Then \(a^2 = \dfrac{16x^4}{49y^2}\), \(b^2 = \dfrac{49y^4}{16x^2}\), and \(2ab = 2\left(\dfrac{4x^2}{7y}\right)\left(\dfrac{7y^2}{4x}\right) = 2xy\) — all three terms match.

\[\frac{16x^4}{49y^2} + \frac{49y^4}{16x^2} + 2xy = \left(\frac{4x^2}{7y} + \frac{7y^2}{4x}\right)^2\]

Expand: \(x^2 + xy – xz – yz\)

Group: \((x^2+xy) – (xz+yz)\)

\[= x(x+y) – z(x+y)\]

\[= (x+y)(x-z)\]

Take out common factor \(8\): \(8(4x^2 – y^2)\)

Apply \(a^2-b^2=(a+b)(a-b)\) with \(a=2x,\ b=y\):

\[32x^2 – 8y^2 = 8(2x+y)(2x-y)\]

Write as difference of squares: \((a^2b^2)^2 – (4)^2\)

\[= (a^2b^2+4)(a^2b^2-4)\]

\(a^2b^2 – 4\) is again a difference of squares \((ab)^2 – 2^2\):

\[= (a^2b^2+4)(ab+2)(ab-2)\]

Apply \(a^2-b^2=(a+b)(a-b)\) with \(a=(x+y+z),\ b=(x+y-z)\):

\(a+b = 2(x+y)\) and \(a-b = 2z\)

\[(x+y+z)^2-(x+y-z)^2 = 2(x+y)\cdot 2z\]

Common factor \(x\): \(x(x^2-1)\)

\(x^2-1\) is a difference of squares:

\[x^3 – x = x(x+1)(x-1)\]

Apply \(a^2-b^2=(a+b)(a-b)\) with first term \((a+1)\), second term \((b-1)\):

Sum: \((a+1)+(b-1) = a+b\)

Difference: \((a+1)-(b-1) = a-b+2\)

Write \(25 = 5^2\): \((a+4)^2 – 5^2\)

\[= \big[(a+4)+5\big]\big[(a+4)-5\big]\]

Write as \((x^2)^2 – (y^2)^2\): \((x^2+y^2)(x^2-y^2)\)

\(x^2-y^2\) factors further:

\[x^4-y^4 = (x^2+y^2)(x+y)(x-y)\]

Write \(4z^2 = (2z)^2\): \((3x+2y)^2 – (2z)^2\)

\[= \big[(3x+2y)+2z\big]\big[(3x+2y)-2z\big]\]

Check perfect square: \(64x^2=(8x)^2,\ 49=7^2,\ 2\cdot8x\cdot7=112x\) — matches \(a^2-2ab+b^2\).

\[64x^2-112x+49 = (8x-7)^2\]

Group first three terms as a perfect square: \((a^2+2ab+b^2) – c^2\)

\[= (a+b)^2 – c^2\]

\[= (a+b+c)(a+b-c)\]

Take out common factor \(3\): \(3(x^2 – 8x + 16)\)

\(x^2-8x+16\) is a perfect square \((x-4)^2\):

\[3x^2-24x+48 = 3(x-4)^2\]

Expand: \(x^2+y^2+2xy+2yz+2zx\)

Group first three as \((x+y)^2\), remaining as \(2z(x+y)\):

\[= (x+y)^2 + 2z(x+y)\]

\[= (x+y)\big[(x+y)+2z\big]\]

Check against \((a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\) with \(a=2x,\ b=-3y,\ c=z\):

\(a^2=4x^2,\ b^2=9y^2,\ c^2=z^2\) — matches.

\(2ab=-12xy,\ 2bc=-6yz,\ 2ca=4xz\) — all match.

Matches the identity \(a^3 – 3a^2b + 3ab^2 – b^3 = (a-b)^3\) with \(a=x,\ b=1\):

Split the middle term: \((bc+ad)x = bcx + adx\)

\[acx^2 + bcx + adx + bd\]

Group: \(cx(ax+b) + d(ax+b)\)

Check against \((a+b+c)^2\) with \(a=3x,\ b=y,\ c=5z\):

\(a^2=9x^2,\ b^2=y^2,\ c^2=25z^2\) — matches.

\(2ab=6xy,\ 2bc=10yz,\ 2ca=30xz\) — all match.

Check against \((a+b+c)^2\) with \(a=7m,\ b=-2n,\ c=-3z\):

\(a^2=49m^2,\ b^2=4n^2,\ c^2=9z^2\) — matches.

\(2ab=-28mn,\ 2bc=12nz,\ 2ca=-42mz\) — all match.

Check against \((a+b+c)^2\) with \(a=p,\ b=\dfrac{q}{2},\ c=1\):

\(a^2=p^2,\ b^2=\dfrac{q^2}{4},\ c^2=1\) — matches.

\(2ab=pq,\ 2bc=q,\ 2ca=2p\) — all match.