$1\text{}({x}^{(1/3)}+{y}^{(1/3)})$

pliwraih
2022-07-28
Answered

Calculate.

$1\text{}({x}^{(1/3)}+{y}^{(1/3)})$

$1\text{}({x}^{(1/3)}+{y}^{(1/3)})$

You can still ask an expert for help

Cael Cox

Answered 2022-07-29
Author has **11** answers

$1/({x}^{(1/3)}+{y}^{(1/3)})$

$=({x}^{(2/3)}+{y}^{(2/3)}-(xy{)}^{(1/3)})/(x+y)$

ie. multiply and divide by $({x}^{(2/3)}+{y}^{(2/3)}-(xy{)}^{(1/3)})$.

$=({x}^{(2/3)}+{y}^{(2/3)}-(xy{)}^{(1/3)})/(x+y)$

ie. multiply and divide by $({x}^{(2/3)}+{y}^{(2/3)}-(xy{)}^{(1/3)})$.

Nash Frank

Answered 2022-07-30
Author has **10** answers

I assume you are required to rationalize the denominator.

Use the sum of cubes formula,${x}^{3}+{y}^{3}=(x+y)({x}^{2}-xy+{y}^{2})$

for your question use $x+y=({x}^{1/3}+{y}^{1/3})({x}^{2/3}-{x}^{1/3}{y}^{1/3}+{y}^{2/3})$

multiply numerator and denominator by $({x}^{2/3}-{x}^{1/3}{y}^{1/3}+{y}^{2/3})$

to get $({x}^{2/3}-{x}^{1/3}{y}^{1/3}+{y}^{2/3})/(x+y)$

Use the sum of cubes formula,${x}^{3}+{y}^{3}=(x+y)({x}^{2}-xy+{y}^{2})$

for your question use $x+y=({x}^{1/3}+{y}^{1/3})({x}^{2/3}-{x}^{1/3}{y}^{1/3}+{y}^{2/3})$

multiply numerator and denominator by $({x}^{2/3}-{x}^{1/3}{y}^{1/3}+{y}^{2/3})$

to get $({x}^{2/3}-{x}^{1/3}{y}^{1/3}+{y}^{2/3})/(x+y)$

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