1. **State the problem:** Show that $\left(\sqrt{12}+\sqrt{3}\right)^2=27$. 2. **Recall the formula:** The square of a sum is given by $$(a+b)^2 = a^2 + 2ab + b^2$$ where $a=\sqrt{12}$ and $b=\sqrt{3}$. 3. **Calculate each term:** - $a^2 = (\sqrt{12})^2 = 12$ - $b^2 = (\sqrt{3})^2 = 3$ - $2ab = 2 \times \sqrt{12} \times \sqrt{3} = 2 \times \sqrt{36} = 2 \times 6 = 12$ 4. **Sum all terms:** $$12 + 12 + 3 = 27$$ 5. **Conclusion:** Therefore, $\left(\sqrt{12}+\sqrt{3}\right)^2 = 27$ as required.