(b⁻⁴)⁻³ ÷ b⁻¹⁷ = b¹² ÷ b⁻¹⁷= b⁻⁵
1/3√900y - √121y = √1/9 × 900y -√121y = √100y - √121y = 10y - 11y = -1y
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(b⁻⁴)⁻³ ÷ b⁻¹⁷ = b¹² ÷ b⁻¹⁷= b⁻⁵
1/3√900y - √121y = √1/9 × 900y -√121y = √100y - √121y = 10y - 11y = -1y