\(1+2+3 = 6.\)The prime factors of 6 are 2 and 3. The greatest one is \(\boxed3\)
I hope this is not too much, but can you solve it with similar triangles?
Expanding the product, we get \(x^7 + 4x^6 + 4x^5 + 19x^4 + 19x^3 + 16x^2 + 16x + 1\). We see that the coefficient of x is 16.
When we expand the product, we get \(x^7 + 4x^6 + 4x^5 + 19x^4 + 16x^2 + 16x + 1.\) We see that the coefficient of \(x^2\) is \(16\).
If we multiply both sides by 3, we get \(3x + 3y = 75\) . We notice that \(6x+3y\) is \(3x\) more than \(3x + 3y\). If \(x\) is 0, then \(3x\) is 0. 75 + 0 = 75, so the minumum value of \(6x + 3y\) is 75.
Anything to the power of 0 is 1, so if x = 0, 9^x would be 1. Lets assume x = 0. Then \((9^x - 2) \cdot (3^x + 1)\) would be equal to \(-1 \cdot 1\) or -1. So, the minimum value of \((9^x - 2) \cdot (3^x + 1)\) is -1.
