Question
Factorise:$2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$
✓
Answer
$2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$$=\big(\sqrt{2}\text{a}\big)^3+\big(\sqrt{3}\text{b}\big)^3+\text{c}^3-3\big(\sqrt{2}\text{a}\big)\big(\sqrt{3}\text{b}\big)\text{c}$
We know
$\text{x}^3+\text{y}^3+\text{z}^3-3\text{xyz}=(\text{x}+\text{y}+\text{z})\\\big(\text{x}^2+\text{y}^2+\text{z}^2-\text{xy}-\text{yz}-\text{zx}\big)$
$\text{x}=\sqrt{2}\text{a},\ \text{y}=\sqrt{3}\text{b},\ \text{z}=\text{c}$
$\big(\sqrt{2}\text{a}\big)^3+\big(\sqrt{3}\text{b}\big)^3+\text{c}^3-3\big(\sqrt{2}\text{a}\big)\big(\sqrt{3}\text{b}\big)\text{c}$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6\text{ab}}-\sqrt{3\text{bc}}-\sqrt{2}\text{ac}\big)$
We know
$\text{x}^3+\text{y}^3+\text{z}^3-3\text{xyz}=(\text{x}+\text{y}+\text{z})\\\big(\text{x}^2+\text{y}^2+\text{z}^2-\text{xy}-\text{yz}-\text{zx}\big)$
$\text{x}=\sqrt{2}\text{a},\ \text{y}=\sqrt{3}\text{b},\ \text{z}=\text{c}$
$\big(\sqrt{2}\text{a}\big)^3+\big(\sqrt{3}\text{b}\big)^3+\text{c}^3-3\big(\sqrt{2}\text{a}\big)\big(\sqrt{3}\text{b}\big)\text{c}$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6\text{ab}}-\sqrt{3\text{bc}}-\sqrt{2}\text{ac}\big)$
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