Question
Factorise:
8 - 27b3 - 343c3 - 126bc
8 - 27b3 - 343c3 - 126bc
✓
Answer
8 - 27b3 - 343c3 - 126bc
= (2)3 + (-3b)3 + (-7c)3 - 3 × (2) × (-3b) × (-7c)
= [2 + (-3b) + (-7c)[(2)2 + (-3b)2 + (-7c)2 - (2)(-3b) - (-3b)(-7c) - (2)(-7c)]
= (2 - 3b - 7c)(4 + 9b2 + 49c2 + 6b - 21bc + 14c)
= (2)3 + (-3b)3 + (-7c)3 - 3 × (2) × (-3b) × (-7c)
= [2 + (-3b) + (-7c)[(2)2 + (-3b)2 + (-7c)2 - (2)(-3b) - (-3b)(-7c) - (2)(-7c)]
= (2 - 3b - 7c)(4 + 9b2 + 49c2 + 6b - 21bc + 14c)
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In Agra in a grinding mill, there were installed 5 types of mills. These mills used steel balls of radius 5mm, 7mm,10mm, 14mm and 16mm respectively. All the balls were in the spherical shape. For repairing purpose mills need 10 balls of 7mm radius and 20 balls of 3.5mm radius. The workshop was having 20000mm3 steel. This 20000mm3 steel was melted and 10 balls of 7mm radius and 20 balls of 3.5mm radius were made and the remaining steel was stored for future use.
questions:
1. What was the volume of 10 balls of radius 7mm?
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In Agra in a grinding mill, there were installed 5 types of mills. These mills used steel balls of radius 5mm, 7mm,10mm, 14mm and 16mm respectively. All the balls were in the spherical shape. For repairing purpose mills need 10 balls of 7mm radius and 20 balls of 3.5mm radius. The workshop was having 20000mm3 steel. This 20000mm3 steel was melted and 10 balls of 7mm radius and 20 balls of 3.5mm radius were made and the remaining steel was stored for future use.
questions:
1. What was the volume of 10 balls of radius 7mm?
2. What was the volume of 20 balls of radius 3.5mm?
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