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प्रश्न
Factorise the following:
27x3 – 8y3
बेरीज
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उत्तर
27x3 – 8y3 = (3x)3 – (2y)3 ...[a3 – b3 = (a – b)(a2 + ab + b2)]
= (3x – 2y)[(3x)2 + 3x × 2y + (2y)3]
= (3x – 2y)(9x3 + 6xy + 4xy + 4y3)
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पाठ 3: Algebra - Exercise 3.5 [पृष्ठ १०५]
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संबंधित प्रश्न
Simplify:
\[\frac{x^2 - 5x - 24}{\left( x + 3 \right)\left( x + 8 \right)} \times \frac{x^2 - 64}{\left( x - 8 \right)^2}\]
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\[\frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}\]
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`(1 - 2x + x^2)/(1 - x^3) xx (1 + x + x^2)/(1 + x)`
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y3 − 27
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p3 − (p + 1)3
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Factorise: `a^3 - 1/(a^3)`
Simplify: (2x + 3y)3 - (2x - 3y)3
