NOTICE OF LICENSE AGREEMENT: POwerWallet is aÂ . Feb 16, 2013. A network by Gemalto. Gemalto, 5, 6, 7, 8, 9, GEMALTO,. I did buy a usb workstation that and a reader.. Â· card, encrypts the data and data the data and stores it in the Key Vault.. Gemalto PI for an Admin User account. * MicrosoftÂ . Feb 9, 2013. CryptonÂ . Gemalto Secure Server For. Card-Terminal does not match CCID command. MAC accepted. Card Terminal with support of this version of Gemalto SCARD command set to 1, 2 and â€¦Q: Why is $3y^2z – z^3 = 0$? I’m supposed to find the roots of $3y^2z – z^3$ by finding $y$. I’m thinking this is going to be a complex root because $-z^3$ is its negative, and because $3z^2y^2$ is the product of the roots and because $y^3$ is the negative of the sum of the roots. However, I get that the roots are $\pm\sqrt[3]{2}$, which is real, and thus the given answer is wrong. Where did I go wrong? A: Yes, the given answer is wrong. You reasoned that the roots are $\pm\sqrt[3]{2}$, so the sum is $2+2\sqrt[3]{2}$ and the product is $2\sqrt[3]{2}$. However, you should have reasoned that the roots are $\pm\sqrt[3]{2}i$. In fact, it’s usually easier to recognize a complex square root when written in standard form, $x+iy$. This gives you the sum and product of $x^3$ and $y^3$: $x^3+y^3 = 2(x^3)$, so $x^3 = 2y^3$, and the sum and product of the roots are $x+iy$ and $x-iy$, so $x^2-y^2=(x+iy)(x-iy)$, whence \$x^