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If a=0 and b=1, find the value of 1. a³+b³+3a²b+3ab²2. a²+2ab+b²3. a²-b²4. (a+b)(a-b)5. (a+b)³6. (a+b)²please follow me I will follow them back |
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Answer» ion ✪If a=0 and b=1, find the VALUE of1. a³+b³+3a²b+3ab²2. a²+2ab+b²3. a²-b²4. (a+b)(a-b)5. (a+b)³6. (a+b)²✪ Given ✪a = 0b = 1✪ To find ✪The values.✪ FORMULAE to be USED ✪a³+b³+3a²b+3ab² = (a+b)³a²+2ab+b² = (a+b)²a²-b² = (a+b)(a-b)✪ Solution ✪1. a³+b³+3a²b+3ab²a³+b³+3a²b+3ab²= (a+b)³= (0+1)³= 1³= 12. a²+2ab+b²a²+2ab+b² = (a+b)²= (0+1)²= 1²= 13. a²-b²a²-b²= 0²-1²= 0-1= -14. (a+b)(a-b)(a+b)(a-b)= a²-b²= 0²-1²= 0-1= -15. (a+b)³(a+b)³= (0+1)³= 1³= 16. (a+b)²(a+b)²= (0+1)²= 1²= 1✪ Hence ✪When a = 0 and b = 1,a³+b³+3a²b+3ab² = 1a²+2ab+b² = 1a²-b² = -1(a+b)(a-b) = -1(a+b)³ = 1(a+b)² = 1________________________________✪ Some more formulae ✪☞ (a+b)² = a²+2ab+b²☞ (a-b)² = a²-2ab+b²☞ a²-b² = (a+b)(a-b)☞ a²+b² = (a+b)²-2ab☞ a²+b² = (a-b)²+2ab☞ 2(a²+b²) = (a+b)²+(a-b)²☞ (a+b)³ = a³+3a²b+3ab²+b³☞ (a-b)³ = a³-3a²b+3ab²-b³☞ a³+b³ = (a+b)(a²-ab+b²)☞ a³-b³ = (a-b)(a²+ab+b²) ๑ Hope this helps you. ๑ |
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