Question
Factorise: $3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
✓
Answer
$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}\big)^3+(-\text{b})^3+\big(-\sqrt{5}\text{c}\big)^3-3\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
We know $\text{x}^3+\text{y}^3+\text{z}^3-3\text{xyz}=(\text{x}+\text{y}+\text{z})\$\text{x}^2+\text{y}^2+\text{z}^2-\text{xy}-\text{yz}-\text{zx})$
Here, $\text{x}=\big(\sqrt{3}\text{a}\big),\ \text{y}=(-\text{b}),\ \text{z}=\big(-\sqrt{5}\text{c}\big)$
$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}\big)^3+(-\text{b})^3+\big(-\sqrt{5}\text{c}\big)^3-3\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
$=\big(\sqrt{3}\text{a}-\text{b}-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{c}\big)$
$=\big(\sqrt{3}\text{a}\big)^3+(-\text{b})^3+\big(-\sqrt{5}\text{c}\big)^3-3\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
We know $\text{x}^3+\text{y}^3+\text{z}^3-3\text{xyz}=(\text{x}+\text{y}+\text{z})\$\text{x}^2+\text{y}^2+\text{z}^2-\text{xy}-\text{yz}-\text{zx})$
Here, $\text{x}=\big(\sqrt{3}\text{a}\big),\ \text{y}=(-\text{b}),\ \text{z}=\big(-\sqrt{5}\text{c}\big)$
$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}\big)^3+(-\text{b})^3+\big(-\sqrt{5}\text{c}\big)^3-3\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
$=\big(\sqrt{3}\text{a}-\text{b}-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{c}\big)$
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