[tex]\frac{a}4=x+3 /4\\ a=4(x+3)\\ a=4x+12\\ \\ \frac{b}4=x+1 /*4\\ b=4(x+1)\\ b=4x+4\\ \\ \frac{a^2-b^2}8=\frac{(4x+12)^2-(4x+4)^2}8=\frac{16x^2+96x+144-(16x^2+32x+16)}8=\frac{16x^2+96x+144-16x^2-32x-16}8=\frac{64x+128}8=\frac{8(8x+16)}8=8x+16[/tex]