Expand (3a+2b)(2a−3b)\Big(3a + \dfrac{2}{b}\Big)\Big(2a - \dfrac{3}{b}\Big)(3a+b2)(2a−b3)
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Given,
(3a+2b)(2a−3b)\Big(3a + \dfrac{2}{b}\Big)\Big(2a - \dfrac{3}{b}\Big)(3a+b2)(2a−b3)
Expanding,
⇒3a×2a+3a×−3b+2b×2a+2b×−3b⇒6a2−9ab+4ab−6b2⇒6a2−5ab−6b2.\Rightarrow 3a \times 2a + 3a \times -\dfrac{3}{b} + \dfrac{2}{b} \times 2a + \dfrac{2}{b} \times -\dfrac{3}{b} \\[1em] \Rightarrow 6a^2 - \dfrac{9a}{b} + \dfrac{4a}{b} - \dfrac{6}{b^2} \\[1em] \Rightarrow 6a^2 - \dfrac{5a}{b} - \dfrac{6}{b^2}.⇒3a×2a+3a×−b3+b2×2a+b2×−b3⇒6a2−b9a+b4a−b26⇒6a2−b5a−b26.
Hence, (3a+2b)(2a−3b)=6a2−5ab−6b2.\Big(3a + \dfrac{2}{b}\Big)\Big(2a - \dfrac{3}{b}\Big) = 6a^2 - \dfrac{5a}{b} - \dfrac{6}{b^2}.(3a+b2)(2a−b3)=6a2−b5a−b26.
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