2 sqrt[3]{40} - 4 sqrt[3]{320} + 3 sqrt[3]{62 | vedclass

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Question

(B) Given expression: $2 \sqrt[3]{40} - 4 \sqrt[3]{320} + 3 \sqrt[3]{625} - 3 \sqrt[3]{5}$Step $1$: Simplify each radical term.$2 \sqrt[3]{40} = 2 \sq

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(B) Given expression: $2 \sqrt[3]{40} - 4 \sqrt[3]{320} + 3 \sqrt[3]{625} - 3 \sqrt[3]{5}$Step $1$: Simplify each radical term.$2 \sqrt[3]{40} = 2 \sqrt[3]{8 	imes 5} = 2 	imes 2 \sqrt[3]{5} = 4 \sqrt[3]{5}$$4 \sqrt[3]{320} = 4 \sqrt[3]{64 	imes 5} = 4 	imes 4 \sqrt[3]{5} = 16 \sqrt[3]{5}$$3 \sqrt[3]{625} = 3 \sqrt[3]{125 	imes 5} = 3 	imes 5 \sqrt[3]{5} = 15 \sqrt[3]{5}$Step $2$: Substitute these values back into the expression.$= 4 \sqrt[3]{5} - 16 \sqrt[3]{5} + 15 \sqrt[3]{5} - 3 \sqrt[3]{5}$Step $3$: Combine the terms.$= (4 - 16 + 15 - 3) \sqrt[3]{5}$$= (19 - 19) \sqrt[3]{5}$$= 0 	imes \sqrt[3]{5} = 0$

$2 \sqrt[3]{40} - 4 \sqrt[3]{320} + 3 \sqrt[3]{625} - 3 \sqrt[3]{5}$ is equal to

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