Check sibling questions

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Transcript

Misc 1 Evaluate: (𝑖^18+(1/i)^25 )^3 (𝑖^18+(1/𝑖)^25 )^3 = (𝑖^18+ 1/(π’Š)^πŸπŸ“ )^3 = (𝑖^18+ 1/(π’Š Γ— π’Š^πŸπŸ’ ))^3 = ((π’Š^𝟐 )^πŸ—+1/(𝑖 Γ— (π’Š^𝟐 )^𝟏𝟐 ))^3 Putting i2 = βˆ’πŸ = ((βˆ’πŸ)^πŸ—+1/〖𝑖 Γ— (βˆ’πŸ)γ€—^12 )^3 = (βˆ’πŸ+1/(𝑖 Γ— 𝟏))^3 = (βˆ’1+1/𝑖)^3 Removing π’Š from the denominator = (βˆ’1+1/π‘–Γ—π’Š/π’Š)^3 = (βˆ’1+𝑖/π’Š^𝟐 )^3 = (βˆ’1+𝑖/((βˆ’πŸ)))^3 = (βˆ’πŸ – π’Š )πŸ‘ = (βˆ’1(1+ 𝑖 ))3 = (βˆ’πŸ)πŸ‘ (𝟏 + π’Š )πŸ‘ = (βˆ’1)(1 + 𝑖 )3 = βˆ’(𝟏 + π’Š )πŸ‘ Using (a + b) 3 = a3 + b3 + 3ab(a + b) = βˆ’(13 + 𝑖3 + 3 Γ— 1 Γ— 𝑖 (1 + 𝑖)) = βˆ’(1 + π’ŠπŸ‘ +3𝑖 (1 + 𝑖)) = βˆ’(1 + π’ŠπŸ Γ— π’Š +3𝑖 (1 + 𝑖)) Putting i2 = βˆ’πŸ = βˆ’(1 +(βˆ’πŸ) Γ— 𝑖 +3𝑖 (1 + 𝑖)) = βˆ’(1 βˆ’π‘– +3𝑖 (1 + 𝑖)) = βˆ’(1 βˆ’π‘– +3𝑖+3𝑖 Γ— 𝑖) = βˆ’(1+2𝑖+3π’Š^𝟐 ) Putting i2 = βˆ’πŸ = βˆ’(1+2𝑖+3 Γ— βˆ’πŸ) = βˆ’(1+2π‘–βˆ’3) = βˆ’(2π‘–βˆ’2) = βˆ’2𝑖+2 = πŸβˆ’πŸπ’Š

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.