Expand Expression: Expand the expression using the binomial theorem or by multiplying (sqrt(5)-3) by itself three times. (sqrt(5)-3)^3 = (sqrt(5)-3) * (sqrt(5)-3) * (sqrt(5)-3)Multiply First Two Factors: Multiply the first two factors (sqrt(5)-3) and (sqrt(5)-3). (sqrt(5)-3) * (sqrt(5)-3) = (sqrt(5))^2 - 2*3*sqrt(5) + 3^2 = 5 - 6*sqrt(5) + 9 = 14 - 6*sqrt(5)Multiply Result with Remaining Factor: Multiply the result from Step 2 by the remaining factor (sqrt(5)-3). (14 - 6*sqrt(5)) * (sqrt(5)-3) = 14*sqrt(5) - 42 - 6*sqrt(5)*sqrt(5) + 18*sqrt(5) = 14*sqrt(5) - 42 - 6*5 + 18*sqrt(5) = 14*sqrt(5) + 18*sqrt(5) - 42 - 30 = (14 + 18)*sqrt(5) - 72 = 32*sqrt(5) - 72

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$(\sqrt{5}-3)^3$

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Grade 7

Powers with negative bases

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(\sqrt{5}-3)^3

Expand Expression: Expand the expression using the binomial theorem or by multiplying $(\sqrt{5}-3)$ by itself three times. $\newline$ $(\sqrt{5}-3)^3 = (\sqrt{5}-3) \times (\sqrt{5}-3) \times (\sqrt{5}-3)$
Multiply First Two Factors: Multiply the first two factors $(\sqrt{5}-3)$ and $(\sqrt{5}-3)$ . $(\sqrt{5}-3) \times (\sqrt{5}-3) = (\sqrt{5})^2 - 2\times3\times\sqrt{5} + 3^2 = 5 - 6\times\sqrt{5} + 9 = 14 - 6\times\sqrt{5}$
Multiply Result with Remaining Factor: Multiply the result from Step $2$ by the remaining factor $(\sqrt{5}-3)$ . $(14 - 6\sqrt{5}) \times (\sqrt{5}-3)$ $= 14\sqrt{5} - 42 - 6\sqrt{5}\sqrt{5} + 18\sqrt{5}$ $= 14\sqrt{5} - 42 - 6\times 5 + 18\sqrt{5}$ $= 14\sqrt{5} + 18\sqrt{5} - 42 - 30$ $= (14 + 18)\sqrt{5} - 72$ $= 32\sqrt{5} - 72$