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Show that x3 + y3 = (x + y )(x2 – xy + y2). Solution: Since (x + y)3 = x3 + y3 + 3xy(x + y)∴ x3 + y3 = [(x + y)3] –3xy(x + y)= [(x + y)(x + y)2] – 3xy(x + y)= (x + y)[(x + y)2 – 3xy]= (x + y)[(x2 + y2 + 2xy) – 3xy]= (x + y)[x2 + y2 – xy]= (x + y)[x2 + y2 – xy]Thus, x3 + y3 = (x + y)(x2 – xy + y2) How? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Show that x3 + y3 = (x + y )(x2 – xy + y2). Solution: Since (x + y)3 = x3 + y3 + 3xy(x + y)∴ x3 + y3 = [(x + y)3] –3xy(x + y)= [(x + y)(x + y)2] – 3xy(x + y)= (x + y)[(x + y)2 – 3xy]= (x + y)[(x2 + y2 + 2xy) – 3xy]= (x + y)[x2 + y2 – xy]= (x + y)[x2 + y2 – xy]Thus, x3 + y3 = (x + y)(x2 – xy + y2) How? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Show that x3 + y3 = (x + y )(x2 – xy + y2). Solution: Since (x + y)3 = x3 + y3 + 3xy(x + y)∴ x3 + y3 = [(x + y)3] –3xy(x + y)= [(x + y)(x + y)2] – 3xy(x + y)= (x + y)[(x + y)2 – 3xy]= (x + y)[(x2 + y2 + 2xy) – 3xy]= (x + y)[x2 + y2 – xy]= (x + y)[x2 + y2 – xy]Thus, x3 + y3 = (x + y)(x2 – xy + y2) How?.

Solutions for Show that x3 + y3 = (x + y )(x2 – xy + y2). Solution: Since (x + y)3 = x3 + y3 + 3xy(x + y)∴ x3 + y3 = [(x + y)3] –3xy(x + y)= [(x + y)(x + y)2] – 3xy(x + y)= (x + y)[(x + y)2 – 3xy]= (x + y)[(x2 + y2 + 2xy) – 3xy]= (x + y)[x2 + y2 – xy]= (x + y)[x2 + y2 – xy]Thus, x3 + y3 = (x + y)(x2 – xy + y2) How? in English & in Hindi are available as part of our courses for Class 9. Download more important topics, notes, lectures and mock test series for Class 9 Exam by signing up for free.

Here you can find the meaning of Show that x3 + y3 = (x + y )(x2 – xy + y2). Solution: Since (x + y)3 = x3 + y3 + 3xy(x + y)∴ x3 + y3 = [(x + y)3] –3xy(x + y)= [(x + y)(x + y)2] – 3xy(x + y)= (x + y)[(x + y)2 – 3xy]= (x + y)[(x2 + y2 + 2xy) – 3xy]= (x + y)[x2 + y2 – xy]= (x + y)[x2 + y2 – xy]Thus, x3 + y3 = (x + y)(x2 – xy + y2) How? defined & explained in the simplest way possible. Besides giving the explanation of Show that x3 + y3 = (x + y )(x2 – xy + y2). Solution: Since (x + y)3 = x3 + y3 + 3xy(x + y)∴ x3 + y3 = [(x + y)3] –3xy(x + y)= [(x + y)(x + y)2] – 3xy(x + y)= (x + y)[(x + y)2 – 3xy]= (x + y)[(x2 + y2 + 2xy) – 3xy]= (x + y)[x2 + y2 – xy]= (x + y)[x2 + y2 – xy]Thus, x3 + y3 = (x + y)(x2 – xy + y2) How?, a detailed solution for Show that x3 + y3 = (x + y )(x2 – xy + y2). Solution: Since (x + y)3 = x3 + y3 + 3xy(x + y)∴ x3 + y3 = [(x + y)3] –3xy(x + y)= [(x + y)(x + y)2] – 3xy(x + y)= (x + y)[(x + y)2 – 3xy]= (x + y)[(x2 + y2 + 2xy) – 3xy]= (x + y)[x2 + y2 – xy]= (x + y)[x2 + y2 – xy]Thus, x3 + y3 = (x + y)(x2 – xy + y2) How? has been provided alongside types of Show that x3 + y3 = (x + y )(x2 – xy + y2). Solution: Since (x + y)3 = x3 + y3 + 3xy(x + y)∴ x3 + y3 = [(x + y)3] –3xy(x + y)= [(x + y)(x + y)2] – 3xy(x + y)= (x + y)[(x + y)2 – 3xy]= (x + y)[(x2 + y2 + 2xy) – 3xy]= (x + y)[x2 + y2 – xy]= (x + y)[x2 + y2 – xy]Thus, x3 + y3 = (x + y)(x2 – xy + y2) How? theory, EduRev gives you an ample number of questions to practice Show that x3 + y3 = (x + y )(x2 – xy + y2). Solution: Since (x + y)3 = x3 + y3 + 3xy(x + y)∴ x3 + y3 = [(x + y)3] –3xy(x + y)= [(x + y)(x + y)2] – 3xy(x + y)= (x + y)[(x + y)2 – 3xy]= (x + y)[(x2 + y2 + 2xy) – 3xy]= (x + y)[x2 + y2 – xy]= (x + y)[x2 + y2 – xy]Thus, x3 + y3 = (x + y)(x2 – xy + y2) How? tests, examples and also practice Class 9 tests.