{"@context":"https://schema.org/","@type":"Quiz","name":"

Real part of $(2-3i)^{6}$ is

","hasPart":{"@type":"Question","typicalAgeRange":"12-22","educationalAlignment":[{"@type":"AlignmentObject","alignmentType":"educationalSubject","targetName":"NET"},{"@type":"AlignmentObject","alignmentType":"educationalLevel","targetName":"NET","educationalFramework":"Common Core","targetUrl":"https://classnotes.xyz/mcqs/real-part-of-2-3i-6-is"}],"about":{"@type":"Thing","name":"NET, NUST Entry Test, Mathematics, Number System, Quiz"},"educationalLevel":"intermediate","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","assesses":["NET","NUST Entry Test","Mathematics","Number System","Quiz"],"name":"

Real part of $(2-3i)^{6}$ is

","comment":{"@type":"Comment","text":"

$(2-3i)^{6}$

$=[(2-3i)^{2}]^{3}$

$=(4+9i^{2}-12i)^{3}$ $\\because (a-b)^{2}=a^{2}+b^{2}-2ab$

$=(4+9(-1)-12i)^{3}$ $\\because i^{2}=-1$

$=(4-9-12i)^{3}$

$=(-5-12i)^{3}$

$=(-1)^{3}(5+12i)^{3}$

$=-[125+1728i^{3}+3(5)(12i)(5+12i)]$ $\\because (a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$

$=-[125+1728(-i)+180i(5+12i)]$ $\\because i^{3}=i^{2}\\cdot i =-i$

$=-[125-1728i+900i+2160i^{2}]$

$=-[125-828i+2160(-1)]$

$=-[125-2160-828i]$

$=-[-2035-828i]$

$=2035+828i$

After the expansion of $(2-3i)^{6}$, we get $2035 + 828i$. Therefore, the real part of the expression will be $2035$.

"},"encodingFormat":"text/markdown","text":"

Real part of $(2-3i)^{6}$ is

","suggestedAnswer":[{"@type":"Answer","position":0,"encodingFormat":"text/html","text":"

$64$

"},{"@type":"Answer","position":2,"encodingFormat":"text/html","text":"

$828$

"},{"@type":"Answer","position":3,"encodingFormat":"text/html","text":"

$132$

"}],"acceptedAnswer":{"@type":"Answer","position":1,"encodingFormat":"text/html","text":"

$2035$

","comment":{"@type":"Comment","text":"

$(2-3i)^{6}$

$=[(2-3i)^{2}]^{3}$

$=(4+9i^{2}-12i)^{3}$ $\\because (a-b)^{2}=a^{2}+b^{2}-2ab$

$=(4+9(-1)-12i)^{3}$ $\\because i^{2}=-1$

$=(4-9-12i)^{3}$

$=(-5-12i)^{3}$

$=(-1)^{3}(5+12i)^{3}$

$=-[125+1728i^{3}+3(5)(12i)(5+12i)]$ $\\because (a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$

$=-[125+1728(-i)+180i(5+12i)]$ $\\because i^{3}=i^{2}\\cdot i =-i$

$=-[125-1728i+900i+2160i^{2}]$

$=-[125-828i+2160(-1)]$

$=-[125-2160-828i]$

$=-[-2035-828i]$

$=2035+828i$

After the expansion of $(2-3i)^{6}$, we get $2035 + 828i$. Therefore, the real part of the expression will be $2035$.

"},"answerExplanation":{"@type":"Comment","text":"

$(2-3i)^{6}$

$=[(2-3i)^{2}]^{3}$

$=(4+9i^{2}-12i)^{3}$ $\\because (a-b)^{2}=a^{2}+b^{2}-2ab$

$=(4+9(-1)-12i)^{3}$ $\\because i^{2}=-1$

$=(4-9-12i)^{3}$

$=(-5-12i)^{3}$

$=(-1)^{3}(5+12i)^{3}$

$=-[125+1728i^{3}+3(5)(12i)(5+12i)]$ $\\because (a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$

$=-[125+1728(-i)+180i(5+12i)]$ $\\because i^{3}=i^{2}\\cdot i =-i$

$=-[125-1728i+900i+2160i^{2}]$

$=-[125-828i+2160(-1)]$

$=-[125-2160-828i]$

$=-[-2035-828i]$

$=2035+828i$

After the expansion of $(2-3i)^{6}$, we get $2035 + 828i$. Therefore, the real part of the expression will be $2035$.

"}}},"url":"https://classnotes.xyz/mcqs/real-part-of-2-3i-6-is"}